Reaction Stoichiometry, Gas Laws, Percent Composition 1) Determine the limiting
ID: 573821 • Letter: R
Question
Reaction Stoichiometry, Gas Laws, Percent Composition
1) Determine 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 g/mol, N2H4 = 32.05 g/mol.
N2O4(l) + 2 N2H4(l) 3 N2(g) + 4 H2O(g)
a) Limiting Reactant_______________
b) Theoretical yield of N2_________________
c)If 4.30 grams of N2 was actually produced determine the percent yield and the grams of each reactant which remain:
Percent yield _____________________
Mass of each reactant remaining ___________________
If the above reaction took place in a sealed container, which of the gaseous products shown would have the higher mole fraction in the gas phase above the liquid solution?_________
e) What volume would 4.30 g of Nitrogen gas occupy at STP? _________________
Explanation / Answer
Moles of N2O4 = mass/molecular weight
= 24g / 92.02g/mol
= 0.261 mol
Moles of N2H4 = mass/molecular weight
= 23.5g / 32.05g/mol
= 0.733 mol
From the stoichiometry of the reaction
1 mol of N2O4 reacts with = 2 mol of N2H4
0.261 mol of N2O4 reacts with = 2*0.261 = 0.522 mol of N2H4
N2O4 is limiting reactant
N2H4 is excess reactant
N2 produced = 0.261 mol N2O4 x 3 mol N2/1 mol N2O4
Theoretical yield of N2 = 0.783 mol x 28g/mol = 21.924 g
Actual yield of N2 = 4.30 g
% yield = actual yield x 100/theoretical yield
= 4.3*100/21.924
= 19.61%
Mass of excess reactant remain = initial - reacted
= 0.733 - 0.522 = 0.211 mol x 32.05 g/mol = 6.762 g
At STP
Temperature T = 273.15 K
Pressure P = 1 atm
Moles of N2 = 4.30 g / 28g/mol = 0.153 mol
Gas constant R = 0.0821 L-atm/mol-K
V = nRT/P
= 0.153 x 0.0821 x 273.15 / 1 atm
= 3.43 L
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.