Answer:
The percentage yield of sodium carbonate is 91.47%
Explanation:
we start by writing the reaction equation:
2NaCl + CaCO3 ——-> Na2CO3 + CaCl2
From the reaction we can see that 2 moles of sodium chloride produced 1 mole of sodium carbonate
Let us calculate the actual number of moles of sodium chloride produced from 112 g of it
Mathematically,
number of moles = mass/molar mass
Molar mass of sodium chloride is 23 + 35.5 = 58.5 g/mole
So the number of moles of sodium chloride produced will be 112/58.5 = 1.91 moles
The number of moles of sodium carbonate produced is half of this = 1.91/2 = 0.955
The mass of sodium carbonate produced from 0.955 moles of it will be;
number of moles * molar mass
The molar mass of sodium carbonate is 106 g/mol
So the number of moles is = 0.955 * 106 = 101.23 g
Mathematically;
percentage yield = actual yield/theoretical yield * 100%
Percentage yield = 92.6/101.23 * 100% = 91.47%