Question

2cs(s) + br₂(g) → 2csbr(s). in an experiment, 1 mole of cs was used and it was the limiting reactant, and the student obtained 0.75moles of csbr. what is the percent yield? (no need to convert to grams; can calculated directly from mols yield) 1) 50% 2) 75% 3) 100% 4) 1 %

Explanation:

Step1: Find theoretical yield

From reaction \(2\text{Cs}(s) + \text{Br}_2(g)
ightarrow 2\text{CsBr}(s)\), 2 moles Cs produce 2 moles CsBr. So 1 mole Cs (limiting) should produce 1 mole CsBr (theoretical yield, \(TY = 1\space mol\)).

Step2: Calculate percent yield

Percent yield formula: \(\%\text{yield} = \frac{\text{Actual Yield}(AY)}{\text{Theoretical Yield}(TY)} \times 100\%\). Given \(AY = 0.75\space mol\), \(TY = 1\space mol\).
\(\%\text{yield} = \frac{0.75}{1} \times 100\% = 75\%\).

Answer:

1. 75%