# A mixture of hydrazine and hydrogen peroxide is used as a fuel for rocket engines. How many grams of hydrazine are needed to react with 0.453 moles of hydrogen peroxide?

## The reaction is $\text{N"_2"H"_4(l)+2"H"_2"O"_2(l) -> "N"_2(g)+4"H"_2"O} \left(g\right)$

Jul 29, 2017

${\text{7.26 g N"_2"H}}_{4}$

#### Explanation:

For starters, you know by looking at the balanced chemical equation

${\text{N"_ 2"H"_ (4(l)) + 2"H"_ 2"O"_ (2(l)) -> "N"_ (2(g)) + 4"H"_ 2"O}}_{\left(g\right)}$

that every mole of hydrazine that takes part in the reaction consumes $2$ moles of hydrogen peroxide. In other words, the two reactants take part in the reaction in a $1 : 2$ mole ratio.

You already know that the reaction consumed $0.453$ moles of hydrogen peroxide, so use this mole ratio to figure out how many moles of hydrazine reacted

0.453 color(red)(cancel(color(black)("moles H"_2"O"_2))) * ("1 mole N"_2"H"_4)/(2color(red)(cancel(color(black)("moles H"_2"O"_2)))) = "0.2265 moles N"_2"H"_4

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

$0.2265 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles N"_2"H"_4))) * "32.045 g"/(1color(red)(cancel(color(black)("mole N"_2"H"_4)))) = color(darkgreen)(ul(color(black)("7.26 g}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the number of moles of hydrogen peroxide.