# Question 73554

Jun 9, 2017

$\text{20 g H"_2"O}$

#### Explanation:

Start by taking a look at the balanced chemical equation that describes this reaction.

As you can see, every $2$ moles of hydrogen gas that take part in the reaction consume $1$ mole of oxygen gas and produce $2$ moles of water.

Use the molar masses of the two reactants to convert the masses to moles

4 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.0color(red)(cancel(color(black)("g")))) = "2 moles H"_2

20 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.625 moles O"_2

Now, you know that $2$ moles of hydrogen gas need $1$ mole of oxygen gas in order to take part in the reaction.

Since you don't have enough moles of oxygen gas available, oxygen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of hydrogen gas will get the chance to react.

This means that the reaction will consume $0.625$ moles of oxygen gas and

0.625 color(red)(cancel(color(black)("moles O"_2))) * "2 moles H"_2/(1color(red)(cancel(color(black)("mole O"_2)))) = "1.25 moles H"_2

and produce

0.625 color(red)(cancel(color(black)("moles O"_2))) * ("2 moles H"_2"O")/(1color(red)(cancel(color(black)("mole O"_2)))) = "1.25 moles H"_2"O"#

To convert the number of moles to grams, use the molar mass of water

$1.25 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("20 g}}}}$

The answer must be rounded to one significant figure, the number of sig figs you have for your values.