Question

What is the number of moles of `O_2` that will react with 10.0g of `H_2` to form water in the equation `2H_2 + O_2 -> 2H_2O`?

Answer 1

2.48 moles of `O_2` will react with 10.0g of `H_2` to form water.

Explanation:

For this problem, we should implement the process of stoichiometry. First, convert the grams of `H_2` into moles:

`10.0g H_2xx(1molH_2)/(2.01588 g H_2)`

Then multiply by your mole to mole ratio:

`10.0 g H_2xx(1 mol H_2)/(2.01588 g H_2)xx(1 mol O_2)/(2 mol H_2)`

Cancel out common terms and perform the math to get your answer:

`10.0 cancel(g H_2)xx(1 cancel(mol H_2))/(2.01588 cancel(g H_2))xx(1 mol O_2)/(2 cancel(mol H_2))= 2.48 mol O_2`

The answer has three sig figs because of the given 10.0g.