How many moles of $H_2$ are needed to react with .80 mol of $N_2$ ?

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Answer 1 · 2015-11-12T08:28:58

$"2.4 mol H"_2"$ are needed to react with $"0.80 mol N"_2"$ .

Balanced Equation

$"N"_2("g")+"3H"_2("g")$ $rarr$ $"2NH"_3("g")$

Multiply the $"0.80 mol N"_2"$ times the mole ratio between $"H"_2"$ and $"N"_2"$ so that moles $"N"_2"$ cancel, leaving moles $"H"_2"$ .

$0.80cancel("mol N"_2)xx(3"mol H"_2)/(1cancel("mol N"_2))="2.4 mol H"_2"$

How many moles of H_2H_2 are needed to react with .80 mol of N_2N_2?