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How many grams of H2 are needed to react with 2.40 g of N2?
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Well you need to first write out your balanced reaction first. \[N_{2} +3H_{2} \rightarrow 2NH_{3}\] Find the molar masses of N2 and H2 Convert to moles \[2.40grams~N_{2}*\frac{ mole }{28grams N_{2} } = \frac{ 2.40 }{ 28 } = 0.086 moles~N_{2}\] \[.086~moles~N_{2} * (\frac{ H_{2}}{ 3N_{2} }) = moles~of~H_{2}*(2~grams/mol) = moles~H_{2}\]
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