determined the limiting reactant and the theoretical yield (in g) of nitrogen th

Question

determined the limiting reactant and the theoretical yield (in g) of nitrogen that can be formed from 24.0 g N2O4 and 23.5 g N2H4 molar masses are as follows:  n2o4 92.02  n2h4 32.05

a. limiting reactant:

b. theoretical yield of N2:

c. Determined percent yield and the grams of reactants which remain:  

     percent yield:

     mass of each reactant remaining:

d. if the reaction took place in a sealed container, which of the gaseous products shown would have the igher mole fraction in the gas phase above the liquide solution?

e. What volume 4.3g of nitrogen gas occupy STP?

Explanation / Answer

a.

First the balanced equation

N2O4(l) + 2 N2H4(l) = 3 N2(g) + 4 H2O(g)

mole of N2O4 = 24.0 / 92.02 = 0.2608 mol

mole of N2H4 = 23.5 /32.05 = 0.7332 mol

as each mole of N2O4 require two moles of N2H4 therefore

mole of N2H4 required for 0.2608 mol N2O4 = 0.2608 * 2 = 0.5216 mol

Thus N2H4 is present in excess and hence limiting reactant is N2O4

b.

Mole of N2 = 3 * mole of N2O4 = 3 * 0.2608 = 0.7824 mol

theoretical yield = 0.7824 mol * 28.01 g/mol = 21.92 g

c.

percent yield = (actual yield/ theoretical yield) * 100 = (actual yield / 21.92) *100

Note: As actual yield is not mentioned in the problem I was unable to do above calculation.

mole of N2H4 remaining = 0.7332 mol - 0.5216 mol = 0.2116 mol

mass of N2H4 remainig = 0.2116 * 32.05 = 6.78 g

d.

As we can see from balanced equation for 3 mole of N2 there will be 4 moles of H2O vapor.

Thus, mole fraction of water will be higher

e.

moles in 4.3 g N2 = 4.3 / 28.01 = 0.1535 mol

each mol at STP occupies 22.4 L volume

Thus, volume occupied by 4.3 g N2 at STP = 0.1535 * 22.4 = 3.4 L

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.