Question #73554

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 color(red)(cancel(color(black)("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.