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Q.What is the amount of water produced when 8 g of hydrogen is reacted with 32 g of oxygen?

(a) 2 moles
(b) 1 mole
(c) 3 moles
(d) 0.5 mole
Jharkhand JacJAC Intermediate Board (1st Year) 2026MCQ· 1mImportance★★★★★
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Convert both masses to moles, find the limiting reagent using the balanced equation, then compute moles of water formed from the limiting reagent.

Step 1: Moles of each reactant.

Moles of H2 = mass / molar mass = 8 g / 2 g mol^-1 = 4 mol.

Moles of O2 = mass / molar mass = 32 g / 32 g mol^-1 = 1 mol.

Step 2: Balanced equation.

2H2(g) + O2(g) -> 2H2O(l)

Stoichiometric ratio needed: 2 mol H2 per 1 mol O2.

Step 3: Identify limiting reagent.

Available ratio H2 : O2 = 4 : 1, but only 2 : 1 is required, so H2 is in excess and O2 is the limiting reagent.

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