1. the balanced chemical equation for the reaction of sodium sulfide with chromium(iii) chloride is given…

1. the balanced chemical equation for the reaction of sodium sulfide with chromium(iii) chloride is given below\n3 na₂s + 2 crcl₃ → 6 nacl + cr₂s₃\n1.505 g 2.158 g? g\na. if you start with 1.505 g of na₂s and 2.158 g of crcl₃, what is the theoretical yield of cr₂s₃? show all work to receive credit, answer alone will not be accepted.\nb. if you actually obtain 0.946 g of cr₂s₃, what is the % yield of chromium(iii) sulfide?

1. the balanced chemical equation for the reaction of sodium sulfide with chromium(iii) chloride is given below\n3 na₂s + 2 crcl₃ → 6 nacl + cr₂s₃\n1.505 g 2.158 g? g\na. if you start with 1.505 g of na₂s and 2.158 g of crcl₃, what is the theoretical yield of cr₂s₃? show all work to receive credit, answer alone will not be accepted.\nb. if you actually obtain 0.946 g of cr₂s₃, what is the % yield of chromium(iii) sulfide?

Answer

Explanation:

Step1: Calculate moles of reactants

Molar mass of $Na_2S$: $(2\times22.99)+32.07 = 78.05\ g/mol$. Moles of $Na_2S=\frac{1.505\ g}{78.05\ g/mol}=0.0193\ mol$. Molar mass of $CrCl_3$: $52.00+(3\times35.45)=158.35\ g/mol$. Moles of $CrCl_3=\frac{2.158\ g}{158.35\ g/mol}=0.0136\ mol$.

Step2: Determine limiting reactant

From the balanced equation $3Na_2S + 2CrCl_3\rightarrow6NaCl + Cr_2S_3$, the mole - ratio of $Na_2S$ to $CrCl_3$ is $\frac{3}{2}$. For $0.0193\ mol$ of $Na_2S$, moles of $CrCl_3$ required is $0.0193\ mol\times\frac{2}{3}=0.0129\ mol$. Since we have $0.0136\ mol$ of $CrCl_3$ and we only need $0.0129\ mol$, $Na_2S$ is the limiting reactant.

Step3: Calculate moles of $Cr_2S_3$ produced

The mole - ratio of $Na_2S$ to $Cr_2S_3$ is $\frac{3}{1}$. Moles of $Cr_2S_3$ produced from $0.0193\ mol$ of $Na_2S$ is $0.0193\ mol\times\frac{1}{3}=0.00643\ mol$.

Step4: Calculate theoretical yield of $Cr_2S_3$

Molar mass of $Cr_2S_3$: $(2\times52.00)+(3\times32.07)=200.21\ g/mol$. Theoretical yield of $Cr_2S_3 = 0.00643\ mol\times200.21\ g/mol = 1.287\ g$.

Step5: Calculate percent yield for part b

Percent yield $=\frac{\text{actual yield}}{\text{theoretical yield}}\times100%=\frac{0.946\ g}{1.287\ g}\times100% = 73.5%$

Answer:

a. $1.287\ g$ b. $73.5%$