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. Â
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved. Â
Step 1. Gather all the information in one place with molar masses above the formulas and masses below them. Â
M_r: Â Â Â Â 52.00 Â 80.91 Â Â Â 291.71
        2Cr  +  6HBr ⟶ 2CrBr₃ + 3H₂
Mass/g: Â 15.00 Â Â 15.00 Â
Step 2. Calculate the moles of each reactant Â
 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 × Â
Step 3. Identify the limiting reactant Â
Calculate the moles of CrCl₃ we can obtain from each reactant. Â
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 Hâ‚‚: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 Hâ‚‚ 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 %
