Question

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

`"N"_2"H"_4(l)+2"H"_2"O"_2(l) -> "N"_2(g)+4"H"_2"O"(g)`

Answer 1

`"7.26 g N"_2"H"_4`

Explanation:

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

`"N"_ 2"H"_ (4(l)) + 2"H"_ 2"O"_ (2(l)) -> "N"_ (2(g)) + 4"H"_ 2"O"_ ((g))`

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