Answer:
(a) 18.03 g
(b) 2.105 L
(c) 85.15 %
Step-by-step explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.
M_r: 52.00 80.91 291.71
2Cr + 6HBr ⟶ 2CrBr₃ + 3H₂
Mass/g: 15.00 15.00
Step 2. Calculate the moles of each reactant
Moles of Cr = 15.00 × 1/52.00
Moles of Cr = 0.2885 mol Cr
Moles of HBr = 15.00 × 1/80.91
Moles of HBr = 0.1854 mol HBr ×
Step 3. Identify the limiting reactant
Calculate the moles of CrCl₃ we can obtain from each reactant.
From Cr:
The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr
Moles of CrBr₃ = 0.2885 × 2/2
Moles of CrBr₃ = 0.2885 mol CrCl₃
From HBr:
The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.
Moles of CrBr₃ = 0.1854 × 2/6
Moles of CrBr₃ = 0.061 80 mol CrBr₃
The limiting reactant is HBr because it gives the smaller amount of CrBr₃.
Step 4. Calculate the theoretical yields of CrBr₃ and H₂.
Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1
Theoretical yield of CrBr₃ = 18.03 g CrCl₃
The molar ratio is 3 mol H₂:6 mol HBr
Theoretical yield of H₂ = 0.1854 × 3/6
Theoretical yield of H₂ = 0.092 70 mol H₂
Step 5. Calculate the volume of H₂ at STP
STP is 1 bar and 0 °C.
The molar volume of a gas at STP is 22.71 L.
Volume = 0.092 70 × 22.71/1
Volume = 2.105 L
Step 6. Calculate the percent yield
% Yield = actual yield/theoretical yield × 100 %
Actual yield = 15.35 g
% yield = 15.35/18.03 × 100
% yield = 85.15 %
