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

Md Ariful Islam

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Md Ariful Islam

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For this problem, we should implement the process of . 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.

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