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<em> limiting reactant problem. </em>
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
<em>Step 1</em>. <em>Gather all the information</em> 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
<em>Step 2</em>. Calculate the <em>moles of each reactant</em>
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 ×
<em>Step 3</em>. Identify the<em> limiting reactant</em>
Calculate the moles of CrCl₃ we can obtain from each reactant.
<em>From Cr</em>:
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
<em>From HBr:
</em>
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.
<em>Step 4</em>. Calculate the <em>theoretical yields</em> 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
<em>Step 5</em>. Calculate the <em>volume of H₂</em> 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
<em>Step 6</em>. Calculate the <em>percent yield
</em>
% Yield = actual yield/theoretical yield × 100 %
Actual yield = 15.35 g
% yield = 15.35/18.03 × 100
% yield = <em>85.15 %
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