How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amounts of both?

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equ

Exercises · 1.6

Na_2CO_3

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Concept understanding — Reaction Rate Stoichiometry

Reaction Rate Stoichiometry – From Intuition to Precision

Imagine you are watching a simple reaction:

$2 \text{NO}_2 \rightarrow 2 \text{NO} + \text{O}_2$

As NO₂ disappears, NO appears twice as fast as O₂ appears. Why? Because the balanced equation says: for every 2 molecules of NO₂ that break apart, you get 2 molecules of NO and 1 molecule of O₂. The numbers in front of the species — the stoichiometric coefficients — tell you the relative speeds at which reactants vanish and products appear.

That is the core idea: reaction rate stoichiometry is the relationship between the rates of change of different species in a chemical reaction, dictated by their coefficients in the balanced equation.

The Intuitive Picture

Think of a factory assembly line. The balanced equation is like a recipe:

2 units of raw material A → 2 units of product B + 1 unit of byproduct C

If the line runs steadily, every time 2 units of A are consumed, 2 units of B are produced and 1 unit of C is produced. So the rate at which A disappears must be twice the rate at which C appears. The rate at which B appears equals the rate at which A disappears (both have coefficient 2).

Note

The stoichiometric coefficients are not speeds themselves — they are scaling factors that connect the speeds of different species.

The Precise Statement

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction (often called the rate of the process, $r$ ) is defined as:

r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Here:

$\frac{d[X]}{dt}$ is the instantaneous rate of change of concentration of species X (in mol L⁻¹ s⁻¹).
The minus sign is used for reactants (their concentration decreases with time).
The plus sign is used for products (their concentration increases with time).
Dividing by the coefficient normalises the rate — so $r$ is the same number for every species.

r = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

This single value $r$ is the intrinsic rate of the reaction, independent of which species you measure.

How to Use It (Step by Step)

Suppose you are given:

$2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$

And you measure that $\frac{d[\text{H}_2]}{dt} = -0.040 \text{ M/s}$ (negative because H₂ is being consumed).

Step 1: Write the rate relation:

r = -\frac{1}{2}\frac{d[\text{H}_2]}{dt} = -\frac{1}{1}\frac{d[\text{O}_2]}{dt} = \frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt}

Step 2: Plug in the known value:

r = -\frac{1}{2}(-0.040) = +0.020 \text{ M/s}

Step 3: Find the rate for O₂:

-\frac{d[\text{O}_2]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{O}_2]}{dt} = -0.020 \text{ M/s}

Step 4: Find the rate for H₂O:

$\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{H}_2\text{O}]}{dt} = 2r = 0.040 \text{ M/s}$ …

Why this formula?

Reaction Rate Stoichiometry: Why the Formula Holds

Let’s start with the core idea: In a chemical reaction, the rate at which reactants disappear and products appear is not arbitrary — it is tied directly to the stoichiometric coefficients in the balanced equation.

The Key Formula

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction ( $R$ ) is defined as:

R = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Where:

$[A], [B], [C], [D]$ are concentrations (in mol/L)
$t$ is time
$a, b, c, d$ are stoichiometric coefficients

Why This Formula Holds: The Reasoning

1. The Physical Meaning of Stoichiometric Coefficients

The coefficients tell us the mole ratio in which substances react or are produced. For example:

2H_2 + O_2 \rightarrow 2H_2O

2 moles of $H_2$ react with 1 mole of $O_2$ to produce 2 moles of $H_2O$ .
This means: for every 2 molecules of $H_2$ that disappear, only 1 molecule of $O_2$ disappears, and 2 molecules of $H_2O$ appear.

Key insight: The number of moles changing per unit time is different for each substance, but the reaction event is the same.

2. The Problem with Raw Rates

If we simply wrote:

\text{Rate} = -\frac{d[H_2]}{dt}

This would be twice the rate of disappearance of $O_2$ (since $H_2$ disappears twice as fast). That’s inconsistent — the same reaction shouldn’t have two different numerical rates.

We need a single, unique rate that describes the reaction itself, not just one substance.

3. The Solution: Normalize by Stoichiometric Coefficients

To get a reaction rate that is the same regardless of which substance we track, we divide each substance’s rate of change by its stoichiometric coefficient.

Why division works:

If $A$ disappears at rate $-\frac{d[A]}{dt}$ , and $a$ moles of $A$ are consumed per reaction event, then the number of reaction events per unit time is:

\text{Reaction events per second} = \frac{-\frac{d[A]}{dt}}{a}

Similarly, for product $C$ appearing at rate $+\frac{d[C]}{dt}$ , with $c$ moles produced per event:

\text{Reaction events per second} = \frac{+\frac{d[C]}{dt}}{c}

Since the same reaction is happening, these must be equal. Hence:

-\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

4. The Sign Convention

Reactants decrease over time → $\frac{d[\text{reactant}]}{dt} < 0$ → we add a negative sign to make the rate positive.
Products increase over time → $\frac{d[\text{product}]}{dt} > 0$ → we use a positive sign. …

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The key is to treat the two reactions separately — HCl reacts with $Na_2CO_3$ in a 2:1 mole ratio and with $NaHCO_3$ in a 1:1 ratio. For an equimolar mixture of 1 g total, the required volume of 0.1 M HCl is 157.9 mL.

Why this approach works

When you mix a strong acid like HCl with a carbonate/bicarbonate mixture, two distinct neutralisation reactions occur. The stoichiometry is not the same for both — each mole of $Na_2CO_3$ consumes 2 moles of HCl (because it first forms $HCO_3^-$ , then $H_2CO_3$ ), while each mole of $NaHCO_3$ consumes only 1 mole of HCl. If you miss this difference, you'll get the wrong volume.

The problem gives a total mass of 1 g, but the two compounds are present in equimolar amounts — equal number of moles, not equal mass. That's the crucial starting point.

Step-by-step solution

1. Write the balanced reactions

For $Na_2CO_3$ :

Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2

For $NaHCO_3$ :

NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2

Watch out

A common mistake is to use a 1:1 ratio for $Na_2CO_3$ as well. Remember: carbonate is dibasic — it takes two protons to fully neutralise it.

2. Define the unknown

Let the number of moles of $Na_2CO_3$ = number of moles of $NaHCO_3$ = $x$ (since equimolar).

Molar masses:

$Na_2CO_3$ : $2(23) + 12 + 3(16) = 106$ g/mol
$NaHCO_3$ : $23 + 1 + 12 + 3(16) = 84$ g/mol

Total mass of mixture:

106x + 84x = 190x = 1 \text{ g}

So:

x = \frac{1}{190} \text{ mol}

3. Calculate moles of HCl required

From $Na_2CO_3$ : $2x$ moles of HCl

From $NaHCO_3$ : $x$ moles of HCl

Total HCl needed:

$2x + x = 3x = 3 \times \frac{1}{190} = \frac{3}{190} \text{ mol}$ …

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Here are the most common mistakes students make on this stoichiometry problem, along with the conceptual fix for each.

1. Writing the Wrong Balanced Chemical Equations

The Mistake: Students often write only one generic reaction (e.g., “ $Na_2CO_3 + HCl \rightarrow NaCl + CO_2 + H_2O$ ”) and forget that $NaHCO_3$ reacts differently. They also frequently forget to balance the equations correctly.

Why it happens: Rushing through the problem without checking the acid-base nature of each salt.

How to Avoid:

Write two separate, balanced reactions.
For $Na_2CO_3$ (a carbonate), the reaction with HCl is:

Na_2CO_3 + 2HCl \rightarrow 2NaCl + CO_2 + H_2O

Note: 1 mole of $Na_2CO_3$ requires 2 moles of HCl.

For $NaHCO_3$ (a bicarbonate), the reaction is:

NaHCO_3 + HCl \rightarrow NaCl + CO_2 + H_2O

Note: 1 mole of $NaHCO_3$ requires 1 mole of HCl.

2. Misinterpreting “Equimolar Amounts”

The Mistake: Students assume “equimolar” means equal mass (e.g., 0.5 g each). This leads to incorrect mole calculations.

Why it happens: Confusing “molar” (moles) with “mass” (grams).

How to Avoid:

“Equimolar” means equal number of moles, not equal mass.
Let the number of moles of $Na_2CO_3$ = number of moles of $NaHCO_3$ = $x$ .
Total mass of mixture = $x \times M_{Na_2CO_3} + x \times M_{NaHCO_3} = 1 \text{ g}$ .
Molar masses:
- $M_{Na_2CO_3} = 106 \text{ g/mol}$
- $M_{NaHCO_3} = 84 \text{ g/mol}$
So: $106x + 84x = 190x = 1 \implies x = \frac{1}{190} \text{ mol}$ .

3. Forgetting to Multiply Moles by the Stoichiometric Coefficient

The Mistake: After finding $x$ , students directly use $x$ as the total moles of HCl needed, ignoring that $Na_2CO_3$ consumes 2 moles of HCl per mole.

Why it happens: Not checking the balanced equation for each reactant.

How to Avoid:

Moles of HCl for $Na_2CO_3$ = $2 \times x = 2 \times \frac{1}{190} = \frac{2}{190} \text{ mol}$ .
Moles of HCl for $NaHCO_3$ = $1 \times x = \frac{1}{190} \text{ mol}$ .
Total moles of HCl required = $\frac{2}{190} + \frac{1}{190} = \frac{3}{190} \text{ mol}$ .

4. Incorrect Volume Calculation from Molarity

The Mistake: Using the formula $M = \frac{n}{V}$ but plugging in volume in mL without converting, or using $V = n \times M$ instead of $V = \frac{n}{M}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GUJCET 2025Set 031 markMCQ
$\text{rate} = -6\frac{d[A]}{dt} = -4\frac{d[B]}{dt} = 3\frac{d[C]}{dt} = 4\frac{d[D]}{dt}$
›Reveal solutionSolution
[!TLDR]

The stoichiometric coefficients come out as 2:3:4:3, matching $2A+3B\to4C+3D$ .

Concept

For a reaction $aA+bB\to cC+dD$ , a single rate is defined by
$\text{rate}=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=\frac{1}{c}\frac{d[C]}{dt}=\frac{1}{d}\frac{d[D]}{dt}.$
Solution

Given $\text{rate}=-6\frac{d[A]}{dt}=-4\frac{d[B]}{dt}=3\frac{d[C]}{dt}=4\frac{d[D]}{dt}$ , the individual rates of change are …
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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. The decomposition of NH3 on platinum surface is zero order reaction. What is the rate of production of N2 if K = 2.5 x 10^-4 mol L^-1 S^-1?$
(a) 7.5 x 10^-4 mol L^-1 S^-1
(b) 8.3 x 10^-5 mol L^-1 S^-1
(c) 2.5 x 10^-4 mol L^-1 S^-1
(d) 5 x 10^-4 mol L^-1 S^-1
›Reveal solutionSolution

For a zero-order reaction the rate equals the rate constant k regardless of concentration; stoichiometry then fixes how each product's formation rate relates to that overall rate.

Reaction: 2NH3(g) --Pt--> N2(g) + 3H2(g), zero order, so Rate = k[NH3]^0 = k = 2.5 x 10^-4 mol L^-1 s^-1. By the standard convention, Rate = -(1/2)d[NH3]/dt = +d[N2]/dt = +(1/3)d[H2]/dt = k. So the rate of production of N2 equals k directly (coefficient 1): d …

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GUJCET 2019Set 131 markMCQ
$3A + 2B \rightarrow 5C$
›Reveal solutionSolution

For $3A + 2B \rightarrow 5C$ , divide each rate by its coefficient; reactants get a minus, the product a plus sign.

Concept — rate expression. Rate $= -\dfrac{1}{a}\dfrac{d[A]}{dt}$ for reactants (concentration falling) and $+\dfrac{1}{c}\dfrac{d[C]}{dt}$ for products (concentration rising).

Steps. …
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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. The decomposition of NH3 on the platinum surface is zero order reaction. If K = 2.5 x 10^-4 mol/litre second^-1, what will be the rate of production of H2 in mol/litre second^-1 unit?$
(a) 7.5 x 10^-4
(b) 2.5 x 10^-4
(c) 5.0 x 10^-5
(d) 0.5 x 10^-6
›Reveal solutionSolution

For a zero-order reaction the rate equals the rate constant k regardless of concentration; stoichiometry then fixes how fast each product forms.

The decomposition is 2NH3(g) --Pt--> N2(g) + 3H2(g). Since it is zero order, Rate = k[NH3]^0 = k = 2.5 x 10^-4 mol/litre/second. This rate (by convention, the rate of the reaction as written, i.e. -1/2 d[NH3]/dt) equals d[N2]/dt = (1/3) d[H2]/dt = k. So: …

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. For the reaction 2A + B -> product, -d[A]/dt = K[A]^2[B]. What will be the rate equation for -d[B]/dt?$
(a) K[A][B]^2
(b) K[2A]^2[B]
(c) (1/2) K[A]^2[B]
(d) K[A][B]^(1/2)
›Reveal solutionSolution

Rate = (1/2)(-d[A]/dt) = -d[B]/dt, so -d[B]/dt = (1/2)k[A]^2[B].

For 2A + B -> products, the unique rate of reaction is:

rate = -(1/2) d[A]/dt = -d[B]/dt.

…

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GUJCET 2015Set C1 markMCQ
$X + Y \rightarrow XY$
›Reveal solutionSolution
[!TLDR]

Order in Y = total order - order in X = $3-2 = 1$ , giving rate $= K[X]^2[Y]$ . Answer: (C).

Concept

For a reaction, the overall order is the sum of the powers of the concentration terms in the experimentally-determined rate law (NCERT/CBSE chemical kinetics). Given the total order and the order with respect to one reactant, the order with respect to the other is found by subtraction.

Solution

Total order $= 3$ . …
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GUJCET 2015Set C1 markMCQ
$X \xrightarrow{\text{Step-I}} Y \xrightarrow{\text{Step-II (slow)}} Z$
›Reveal solutionSolution
[!TLDR] The slow (rate-determining) step fixes the observed order; with total order = 2 and Step-II slow, the molecularity of Step-II is 2.

Concept

In a multistep (complex) reaction, the rate-determining step is the slowest step, and it controls the overall rate law. The molecularity of an elementary step is the number of species colliding in that step. For the slow elementary step, the number of reacting molecules (molecularity) corresponds to the experimentally observed overall order of the reaction.

Solution …
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GUJCET 2015Set C1 markMCQ
$3ClO^- \rightarrow ClO_3^- + 2Cl^-$
(i) $ClO^- + ClO^- \xrightarrow{K_1} ClO_2^- + Cl^-$ (Slow step)
(ii) $ClO_2^- + ClO^- \xrightarrow{K_2} ClO_3^- + Cl^-$ (Fast step) then the rate of given reaction = _____. (A) $K_1[ClO^-]$ (B) $K_1[ClO^-]^2$ (C) $K_2[ClO_2^-][ClO^-]$ (D) $K_2[ClO^-]^3$
›Reveal solutionSolution
[!TLDR] The rate equals that of the slow step, $K_1[ClO^-]^2$ , option (B).

Concept

In a reaction mechanism, the overall rate law is fixed by the rate-determining (slowest) elementary step. For an elementary step, the rate is the product of the rate constant and the concentrations of its reactants raised to their stoichiometric coefficients.

Solution

The slow step is

$ClO^- + ClO^- \xrightarrow{K_1} ClO_2^- + Cl^-.$ …
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Q.How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amounts of both?

Concept understanding — Reaction Rate Stoichiometry

Reaction Rate Stoichiometry – From Intuition to Precision

The Intuitive Picture

The Precise Statement

How to Use It (Step by Step)

Why this formula?

Reaction Rate Stoichiometry: Why the Formula Holds

The Key Formula

Why This Formula Holds: The Reasoning

1. The Physical Meaning of Stoichiometric Coefficients

2. The Problem with Raw Rates

3. The Solution: Normalize by Stoichiometric Coefficients

4. The Sign Convention

Why this approach works

Step-by-step solution

1. Write the balanced reactions

2. Define the unknown

3. Calculate moles of HCl required

Method: Acid-Base Stoichiometry for a Carbonate/Bicarbonate Mixture

Step 1 -- Write the balanced reactions

Step 2 -- Set up the equimolar mixture

Step 3 -- Total moles of HCl required …

1. Writing the Wrong Balanced Chemical Equations

2. Misinterpreting “Equimolar Amounts”

3. Forgetting to Multiply Moles by the Stoichiometric Coefficient

4. Incorrect Volume Calculation from Molarity

Concept

Solution

Concept

Solution

Concept

Solution …

Concept

Solution

Q.How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3Na_2CO_3Na2​CO3​ and NaHCO3NaHCO_3NaHCO3​ containing equimolar amounts of both?

Why this approach works

Step-by-step solution

1. Write the balanced reactions

2. Define the unknown

3. Calculate moles of HCl required

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Q.How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amounts of both?

Unlock everything free for 14 days