Question e91cf

Dec 6, 2017

${\text{10 g O}}_{2}$

Explanation:

You know that burning hydrogen gas in oxygen gas will produce water, as shown by the following balanced chemical equation

$2 {\text{H"_ (2(g)) + "O"_ (2(g)) -> 2"H"_ 2"O}}_{\left(l\right)}$

Now, you need to compare the number of moles of hydrogen gas and the number of moles of oxygen gas present in your samples, so use the molar masses of the two reactants to go from grams to moles.

10 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.0color(red)(cancel(color(black)("g")))) = "5.0 moles H"_2

90 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32color(red)(cancel(color(black)("g")))) = "2.8 moles O"_2

According to the balanced chemical equation, every mole of oxygen gas that takes part in the reaction consumes $2$ moles of hydrogen gas.

This means that the number of moles of hydrogen gas present in your sample will only be enough for

5.0 color(red)(cancel(color(black)("moles H"_2))) * "1 mole O"_2/(2color(red)(cancel(color(black)("moles H"_2)))) = "2.5 moles O"_2#

The remaining

$\text{2.8 moles " - " 2.5 moles" = "0.3 moles}$

of oxygen gas are in excess, i.e. they don't take part in the reaction. To convert the number of moles to grams, use the molar mass of oxygen gas.

$0.3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles O"_2))) * "32 g O"_2/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("10 g}}}}$

The answer must be rounded to one significant figure.