Answer:
the mass of C9H8O4 produced from 67.4 g of C7H6O3 is 94.19 g.
Step-by-step explanation:
To solve this problem, we need to use stoichiometry and the given yield percentage. First, we need to write and balance the chemical equation for the reaction:
C7H6O3 (s) + C4H6O3 (s) → C9H8O4 (s) + HC2H3O2 (aq)
From the equation, we can see that one mole of C7H6O3 reacts with one mole of C4H6O3 to produce one mole of C9H8O4. Therefore, we can calculate the number of moles of C9H8O4 produced from 67.4 g of C7H6O3 as follows:
molar mass of C7H6O3 = 122.12 g/mol
moles of C7H6O3 = 67.4 g / 122.12 g/mol = 0.551 moles
Since the reaction has a 95.0% yield, the actual amount of C9H8O4 produced will be 95.0% of the theoretical yield. Therefore, we can calculate the theoretical yield of C9H8O4 as follows:
moles of C9H8O4 = moles of C7H6O3 = 0.551 moles
mass of C9H8O4 = moles of C9H8O4 × molar mass of C9H8O4
mass of C9H8O4 = 0.551 mol × 180.16 g/mol
mass of C9H8O4 = 99.15 g
Finally, we can calculate the actual yield of C9H8O4 as 95.0% of the theoretical yield:
actual yield of C9H8O4 = 95.0% × 99.15 g
actual yield of C9H8O4 = 94.19 g