Question

A mixture of 0.0375 g of h and 0.0185 moles of o in a closed container is sparked to initiate a reaction. How many grams of water can form? Which reactant is in excess? And how many grams of it remain after the reaction?

Answer 1

When 0.0375 g of hydrogen reacts with 0.0185 moles of oxygen, 0.6667 g of water can form. Hydrogen is the reactant in excess, with 0.0002016 g remaining after the reaction.

Step-by-step explanation:

To determine how many grams of water can form from a reaction of hydrogen and oxygen, we use the balanced chemical equation:

2H2(g) + O2(g) → 2H2O(l)

First, we need to calculate the moles of hydrogen from its mass:

0.0375 g H ÷ (1.008 g/mol) = 0.0372 mol H2

Second, since the reaction requires two moles of hydrogen for every mole of oxygen, and we have 0.0185 moles of O2, this means that the hydrogen is in excess:

(2 moles H2 per 1 mole O2) × 0.0185 moles O2 = 0.0370 moles H2 required

Third, to find out how many grams of water can form:

0.0185 moles O2 × (2 moles H2O / 1 mole O2) × (18.02 g/mol H2O) = 0.6667 g H2O

Finally, to calculate the excess hydrogen:

(0.0372 mol H2 - 0.0370 mol H2) × (1.008 g/mol) = 0.0002 mol H2 × 1.008 g/mol = 0.0002016 g remaining H2