Question

es of hydrogen and excess oxygen. 2H2(g)+O2(g)→2H2O(l) Which of the following shows calculations for a correct way to solve this problem? View Available Hint(s) Consider the balanced chemical equation that follows. You are asked to determine how many moles of water you can form from 4 moles of hydrogen and excess oxygen. Which of the following shows calculations for a correct way to solve this problem? 4 mol H2×2 mol H2O2 mol H2=4 mol H2O 4 mol H2×2 mol H2O1 mol O2=8 mol H2O 2 mol H2×2 mol H22 mol H2O=2 mol H2O

Answer 1

Answer: The correct answer is $4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O$

Explanation:

We are given:

Moles of hydrogen gas = 4 moles

As, oxygen is given in excess. Thus, is considered as an excess reagent and hydrogen is considered as a limiting reagent because it limits the formation of products.

For the given chemical equation:

$2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$

By Stoichiometry of the reaction:

2 moles of hydrogen produces 2 moles of water molecule.

So, 4 moles of hydrogen will produce = $\frac{2molH_2O}{2molH_2}\times 4molH_2=4mol$ of water.

Hence, the correct answer is $4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O$