Answer :
The theoretical yield of compound C is approximately 87.1 grams.
Determine the Limiting Reagent:
To find the limiting reagent, we need to compare the number of moles of each reactant with the stoichiometric coefficients in the balanced chemical equation.
Compound A (with a molar mass of m_A = 32.0 g/mol:
[tex]\[n_A = \frac{67.5 g}{m_A} = \frac{67.5 g}{32.0 g/mol} = 2.11 mol\][/tex]
Compound B (with a molar mass of m_B = 18.0 g/mol:
[tex]\[n_B = \frac{39.1 g}{m_B} = \frac{39.1 g}{18.0 g/mol} = 2.17 mol\][/tex]
From the balanced chemical equation, we see that 2 moles of compound A react with 3 moles of compound B to produce 1 mole of compound C.
Comparing the moles available:
- Compound A: 2.11 mol
- Compound B: [tex]\(2.17 mol \times \frac{2 mol}{3 mol} = 1.45 mol \)[/tex]
Since compound B produces fewer moles of compound C, it is the limiting reagent.
2: Calculate Theoretical Yield of Compound C:
Using the stoichiometry from the balanced equation, we can find the theoretical yield of compound C from compound B:
[tex]\[n_C = n_B \times \frac{1 mol}{3 mol} = 1.45 mol \times \frac{1 mol}{3 mol} = 0.483 mol\][/tex]
Now, to find the mass of compound C:
[tex]\[m_C = n_C \times m_C = 0.483 mol \times 180.0 g/mol = 87.1 g\][/tex]
Therefore, the theoretical yield of compound C is approximately 87.1 grams.
