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Ganezh [65]
2 years ago
14

For the reaction n2(g) + 2h2(g) â n2h4(l), if the percent yield for this reaction is 77.5%, what is the actual mass of hydrazine

(n2h4) produced when 27.70 g of nitrogen reacts with 4.45 g of hydrogen?
Chemistry
1 answer:
Rudiy272 years ago
7 0

First calculate the moles of N2 and H2 reacted.

moles N2 = 27.7 g / (28 g/mol) = 0.9893 mol

moles H2 = 4.45 g / (2 g/mol) = 2.225 mol

 

We can see that N2 is the limiting reactant, therefore we base our calculation from that.

Calculating for mass of N2H4 formed:

mass N2H4 = 0.9893 mol N2 * (1 mole N2H4 / 1 mole N2) * 32 g / mol * 0.775

<span>mass N2H4 = 24.53 grams</span>

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In this case, considering the given chemical reaction and the mass of the magnesium strip, following the indications of the atomic weight ratio (2.61 g Cu/1 g Mg), and keeping in mind the 1:1 mole ratio one could compute the produced mass of copper as shown below:

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