Q.In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?

Step-by-step reasoning

1. Balanced equation.

2. Convert to moles.

   Molar mass of CH$_4$ = 16 g/mol, so moles of CH$_4$ = $24/16 = 1.5$ mol. Moles of O$_2$ = 4 mol (given).

3. Find the limiting reagent.

   1.5 mol CH$_4$ requires $1.5 \times 2 = 3$ mol O$_2$. We have 4 mol O$_2$ available — more than enough — so methane is limiting.

4. Moles of CO$_2$ produced.

   1 mol CH$_4$ gives 1 mol CO$_2$, so 1.5 mol CH$_4$ gives 1.5 mol CO$_2$.

5. Convert to molecules.

   $1.5 \times 6.022\times10^{23} = 9.033\times10^{23} \text{ molecules}$ …

Concept and Intuition

When two solutions are mixed, a precipitation reaction occurs if the product is insoluble. Here, silver nitrate and sodium chloride react to form silver chloride, which is a white solid. The mass of precipitate depends entirely on the limiting reagent — the reactant that runs out first. We are given mass percentages, not molarities, so we must first find the actual mass of each solute in the mixed volume, then convert to moles, and finally see which one limits the reaction.

A common mistake is to assume equal volumes mean equal moles, but the concentrations (mass percentages) are different, so the mole amounts differ. Always check the stoichiometry.

Step-by-step solution

1. Find the mass of each solute in the given volumes.

We have 50 mL of each solution. The density of dilute aqueous solutions is approximately $1\ \mathrm{g/mL}$, so 50 mL ≈ 50 g of solution.

• For $\mathrm{AgNO_3}$:

  Mass of solution = 50 g

  Percentage = $16.9\%$

  Mass of $\mathrm{AgNO_3}$ = $\frac{16.9}{100} \times 50 = 8.45\ \mathrm{g}$

• For $\mathrm{NaCl}$:

  Mass of solution = 50 g

  Percentage = $5.8\%$

  Mass of $\mathrm{NaCl}$ = $\frac{5.8}{100} \times 50 = 2.9\ \mathrm{g}$

2. Convert these masses to moles.

Molar mass of $\mathrm{AgNO_3}$:

$\mathrm{Ag} = 107.8,\ \mathrm{N} = 14,\ \mathrm{O_3} = 48$

Total = $107.8 + 14 + 48 = 169.8\ \mathrm{g/mol}$

Moles of $\mathrm{AgNO_3}$ = $\frac{8.45}{169.8} \approx 0.04976\ \mathrm{mol}$

Molar mass of $\mathrm{NaCl}$:

$\mathrm{Na} = 23,\ \mathrm{Cl} = 35.5$

Total = $23 + 35.5 = 58.5\ \mathrm{g/mol}$

Moles of $\mathrm{NaCl}$ = $\frac{2.9}{58.5} \approx 0.04957\ \mathrm{mol}$

3. Write the balanced reaction and identify the limiting reagent.

The mole ratio is 1:1.

We have:

• $\mathrm{AgNO_3}$: $0.04976\ \mathrm{mol}$
• $\mathrm{NaCl}$: $0.04957\ \mathrm{mol}$ …

The balanced reaction is $4\,\text{NH}_3(g)+5\,\text{O}_2(g)\rightarrow 4\,\text{NO}(g)+6\,\text{H}_2\text{O}(g)$, so the stoichiometric ratio is $\text{NH}_3:\text{O}_2 = 4:5$.

Start with $1$ mol $\text{NH}_3$ and $1$ mol $\text{O}_2$.

• To consume all $1$ mol $\text{NH}_3$ we would need $\dfrac{5}{4}\times 1 = 1.25$ mol $\text{O}_2$ — but only $1$ mol is available.
• To consume all $1$ mol $\text{O}_2$ we would need $\dfrac{4}{5}\times 1 = 0.8$ mol $\text{NH}_3$ — and $1$ mol is available (excess). …

Concept & Intuition

This is a classic limiting‑reagent problem disguised as a precipitation reaction. Both reactants are in solution, each with a given concentration (percentage by mass) and volume. We must first convert the percentage to actual mass of solute, then to moles, then use the balanced equation to see which reactant runs out first. The product mass is determined entirely by the limiting reagent.

Step‑by‑Step Solution

1. Write the balanced chemical equation

The mole ratio is 1:1, so the limiting reagent is simply the one with fewer moles.

2. Find the mass of AgNO₃ in the 150 mL of 32% solution “32% solution” means 32 g of AgNO₃ per 100 g of solution. We need the mass of the solution first. Density is not given, so we assume the density of the solution is approximately 1 g/mL (a common approximation for dilute aqueous solutions).

3. Find moles of AgNO₃ Molar mass of AgNO₃ = 108 (Ag) + 14 (N) + 3×16 (O) = 170 g/mol

4. Find the mass of NaCl in the 150 mL of 11% solution Similarly, mass of NaCl solution = 150 g (assuming density ≈ 1 g/mL).

5. Find moles of NaCl Molar mass of NaCl = 23 (Na) + 35.5 (Cl) = 58.5 g/mol

6. Identify the limiting reagent Compare moles:
   • AgNO₃: 0.28235 mol …

Step 1 — HCl produced per day:

Step 2 — Neutralisation stoichiometry:

$n_{\mathrm{Mg(OH)_2}} = \tfrac{1}{2}(0.2055) = 0.1027\ \text{mol}$ …

Balanced reaction:

Moles of $\mathrm{CaCO_3}$ ($M = 100\,\mathrm{g/mol}$):

Moles of HCl required (1 : 2 ratio): …

Step 1 — Why we must treat the two N atoms separately.

Ammonium nitrate is an ionic compound:

One nitrogen sits in the cation (bonded to hydrogen, which is less electronegative than N) and the other in the anion (bonded to oxygen, which is more electronegative than N). So they cannot possibly have the same oxidation number — which immediately kills options (A) and (B).

Step 2 — Nitrogen in the ammonium ion $\text{NH}_4^{+}$.

H in a compound with a non-metal has O.N. $=+1$. The sum of oxidation numbers must equal the ion's charge:

N is more electronegative than H, so it pulls the shared electrons and takes a negative oxidation number. $\checkmark$

Step 3 — Nitrogen in the nitrate ion $\text{NO}_3^{-}$.

O in a normal oxide has O.N. $=-2$:

N is less electronegative than O, so it is oxidised to its maximum, $+5$ (the group number of nitrogen — its highest possible oxidation state). $\checkmark$

Step 4 — Overall check. …

The key idea is simple: a gas is a "greenhouse gas" only if it can absorb the infrared (heat) radiation that the Earth emits. Not every gas in the atmosphere does this. Diatomic molecules like $O_2$ and $N_2$ — which make up most of our air — are transparent to infrared light because their molecular vibrations don't create a changing dipole moment. Without that, they can't trap heat.

Let's check each option:

1. $NO_2$ (Nitrogen dioxide) — This is a triatomic molecule with asymmetric stretching and bending modes. These vibrations change the molecule's electric dipole, so $NO_2$ absorbs infrared radiation. It is indeed a greenhouse gas, though less abundant than $CO_2$.

2. CFC (Chlorofluorocarbon) — These are complex molecules containing carbon, chlorine, and fluorine. They have many vibrational modes that strongly absorb infrared radiation. CFCs are potent greenhouse gases — in fact, a single CFC molecule can trap thousands of times more heat than one $CO_2$ molecule.

3. $CO_2$ (Carbon dioxide) — The classic greenhouse gas. Its linear structure allows asymmetric stretching and bending vibrations that produce a changing dipole moment, making it a strong absorber of infrared radiation. This is the primary driver of anthropogenic climate change.

4. $O_2$ (Oxygen) — A homonuclear diatomic molecule ($O=O$). Its only vibration is a symmetric stretch, which does not change the molecule's dipole moment (since both atoms are identical, the bond is nonpolar). Without a changing dipole, it cannot absorb infrared radiation. It is completely transparent to Earth's heat radiation. …

Concept and Intuition

This is a limiting reagent problem combined with the molar volume of a gas at STP. The reaction is straightforward:

You are given masses of both reactants. One will run out first — that's the limiting reagent — and it determines how much HCl can form. Once you know the moles of HCl produced, you use the fact that at $273$ K and $1$ bar (standard temperature and pressure, STP), one mole of any ideal gas occupies $22.7$ L (not $22.4$ L, because the standard pressure is $1$ bar, not $1$ atm).

Step-by-step solution

1. Write the balanced chemical equation

One mole of $\ce{H2}$ reacts with one mole of $\ce{Cl2}$ to give two moles of $\ce{HCl}$.

2. Find the moles of each reactant

• Molar mass of $\ce{H2}$ = $2.0$ g/mol

  Moles of $\ce{H2}$ = $\frac{0.4}{2.0} = 0.2$ mol

• Molar mass of $\ce{Cl2}$ = $71.0$ g/mol

  Moles of $\ce{Cl2}$ = $\frac{7.1}{71.0} = 0.1$ mol

3. Identify the limiting reagent

From the equation, $1$ mol $\ce{H2}$ needs $1$ mol $\ce{Cl2}$.

We have $0.2$ mol $\ce{H2}$ and $0.1$ mol $\ce{Cl2}$ — chlorine is only half of what would be needed to use up all the hydrogen. So $\ce{Cl2}$ is the limiting reagent. …