Question

Antimony reacts with sulphur according to the equation
\[2Sb\left( s \right) + 3S\left( s \right) \to S{b_2}{S_3}\left( s \right)\]
The molar mass of $S{b_2}{S_3}$ is $340gmo{l^{ - 1}}$. What is the percentage yield for a reaction in which $1 \cdot 40g$ of $S{b_2}{S_3}$ is obtained from $1 \cdot 73g$ of antimony and a slight excess of sulphur$?$
A.$80 \cdot 9\% $
B.$58.0\% $

Answer 1

Hint:In the question above yield percentage has to be calculated, the required parameters are the original or actual yield obtained and calculated or predicted yield which is as whole multiplied by $100$ to obtain the percentage.
Formula used: In order to obtain yield percentage, following formula has to be used $\% yield = \dfrac{{Original}}{{calculated}} \times 100$

Complete step by step answer:
Before finding the yield percentage, we have to note down the molar masses of the following,
Antimony $Sb = 121 \cdot 76g/mol$
Sulphur $S = 32g/mol$
Antimony trisulphide $S{b_2}{S_3} = 340g/mol$
Next, we have to moles of $Sb$ from the given $1 \cdot 73g$ of antimony using which antimony trisulphide is obtained and the molar mass of antimony $Sb = 121.76g/mol$
$1 \cdot 73g \times \left( {\dfrac{{1mol}}{{121 \cdot 76}}} \right) = 0 \cdot 014mol$ $\left( {Sb} \right)$
Now, using the mole ratio we found earlier, the moles of antimony trisulphide $\left( {S{b_2}{S_3}} \right)$is found,
$0 \cdot 014mol\left( {Sb} \right) \times \left( {\dfrac{{1mol\left( {S{b_2}{S_3}} \right)}}{{2mol\left( {Sb} \right)}}} \right) = 0.007mol\left( {S{b_2}{S_3}} \right)$
Now, the moles of antimony trisulphide should be converted to grams $\left( g \right)$
$0 \cdot 007mol\left( {S{b_2}{S_3}} \right) \times \left( {\dfrac{{340g/mol}}{{1mol}}} \right) = 2 \cdot 38grams$
Now, finally calculation the yield percentage,
$\% yield = \left( {\dfrac{{actual}}{{predicted}}} \right) \times 100$
$\% yield = \left( {\dfrac{{1 \cdot 4}}{{2 \cdot 377}}} \right) \times 100 = 58.8\% $
The correct option for the question is (B).

Additional information: The series of steps are written to obtain the percentage of yield. Each and every steps indicate the importance of the substances, units and other values like molar masses and moles. We need to focus on these points in order to understand and make it easy to obtain the required solution.

Note:
The molar masses of substances are required to be known, most the time it is given as in to make the solution a bit easier. Focusing on key points is important as it is the solution that is done step by step and where every step is important. Even the values that have to be calculated should be looked after, because in most of the cases due to small mistakes in signs and mistakes while writing values will make a huge impact on everything, which in turn make us get variations in the final value. General remainder is that looking into units right after the values play a major role, missing those leads to confusion and due to that there will be reduction of marks.