# 3 g of "H"_2 react with 29 g of "O"_2 to give "H"_2"O". Which is the limiting reagent? Calculate the amount of "H"_2"O" formed? Calculate the amount of unreacted reactant left.

Jun 30, 2017

Hydrogen gas will be the limiting reagent.

#### Explanation:

For starters, you need a balanced chemical equation that describes this reaction

So, you know that every mole of oxygen gas that takes part in the reaction consumes $2$ moles of hydrogen gas and produces $2$ moles of water.

The problem provides you with the masses of the two reactants, so use their respective molar mass to convert this to moles

3 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "1.488 moles H"_2

29 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.906 moles O"_2

Now, your goal here is to figure out if you have enough moles of hydrogen gas present to ensure that all the moles of oxygen gas take part in the reaction.

In your case, you can say that $0.906$ moles of oxygen gas would require

0.906 color(red)(cancel(color(black)("moles O"_2))) * "2 moles H"_2/(1color(red)(cancel(color(black)("mole O"_2)))) = "1.812 moles H"_2

SInce you have fewer moles of hydrogen gas than you would need

overbrace("1.488 moles H"_2)^(color(blue)("what you have")) " " < " " overbrace("1.812 moles H"_2)^(color(blue)("what you need"))

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of oxygen will get the chance to react.

This means that the reaction will consume $1.488$ moles of hydrogen gas and

1.488 color(red)(cancel(color(black)("moles H"_2))) * "1 mole O"_2/(2color(red)(cancel(color(black)("moles H"_2)))) = "0.744 moles O"_2

and produce

1.488 color(red)(cancel(color(black)("moles H"_2))) * ("2 moles H"_2"O")/(2color(red)(cancel(color(black)("moles H"_2)))) = "1.488 moles H"_2"O"

To convert this to grams, use the molar mass of water

$1.488 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("27 g}}}}$

I'll leave the answer rounded to two sig figs, but do not forget that you only have one significant figure for the mass of hydrogen gas.

After the reaction is complete, you will be left with

${\text{0.906 moles O"_2 - "0.744 moles O"_2 = "0.162 moles O}}_{2}$

To convert this to grams, use the molar mass of oxygen gas

$0.162 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("5.2 g}}}}$

Once again, I'll leave the answer rounded to two sig figs.