Step-by-step explanation:
28.1 g of Sb₄O₆ are reacting with an excess of C to give 17.3 g of Sb and CO. The unbalanced reaction between these compounds is:
__ Sb₄O₆ + __ C ----> __ Sb + __ CO
We have 4 atoms of Sb on the left and just one on the right. The coefficient for Sb must have 4. Also we have 6 atoms of O on the left and just one on the right side of the equation. So the coefficient for CO must be 6.
__ Sb₄O₆ + __ C ----> 4 Sb + 6 CO
All the elements are balanced except from the C. We have 6 atoms on the right side and just one on the left. The coefficient for C must be 6. The balanced equation is:
Sb₄O₆ + 6 C ----> 4 Sb + 6 CO
To find the percent yield we will have to find the theoretical yield, the mass of Sb that should theoretically be produced when 28.1 g of Sb₄O₆ are reacting with excess of C.
First we have to convert 28.1 g of Sb₄O₆ to moles using the molar mass of Sb₄O₆.
molar mass of Sb = 121.76 g/mol
molar mass of O = 16.00 g/mol
molar mass of Sb₄O₆ = 4 * 121.76 g/mol + 6 * 16.00 g/mol
molar mass of Sb₄O₆ = 583.04 g/mol
mass of Sb₄O₆ = 28.1 g
moles of Sb₄O₆ = 28.1 g * 1 mol/(583.04 g)
moles of Sb₄O₆ = 0.0482 moles
Once we know the moles of Sb₄O₆ we can find the moles of Sb produced.
Sb₄O₆ + 6 C ----> 4 Sb + 6 CO
According to the coefficients of the reaction 1 mol of Sb₄O₆ reacting with an excess of C will produce 4 moles of Sb. Then the molar ratio between Sb₄O₆ and Sb is 1 to 4. We can use that relationship to find the number of moles of Sb that are produced in our exercise.
1 mol of Sb₄O₆ = 4 moles of Sb
moles of Sb = 0.0482 moles of Sb₄O₆ * 4 moles of Sb/(1 mol of Sb₄O₆)
moles of Sb = 0.193 moles
To find the theoretical yield we still have to convert these moles into grams using the molar mass of Sb.
molar mass of Sb = 121.76 g/mol
mass of Sb = 0.193 moles * 121.76 g/mol
mass of Sb = 23.5 g = theoretical yield
And finally we can find the % yield for our reaction.
actual yield = 17.3 g
% yield = actual yield/theoretical yield * 100
% yield = 17.3 g/23.5 g * 100
% yield = 73.6 %
Answer: The percent yield for the reaction is 73.6 %.