A student starts with 25 grams of sodium chloride as the limiting reactant and obtained 17.9 grams of dry pure sodium carbonate.
What is the theoretical yield of sodium carbonate?
What is the percent yield obtained by the student?
To determine the theoretical yield of sodium carbonate, we need to first calculate the stoichiometry of the reaction between sodium chloride (NaCl) and sodium carbonate (Na2CO3). The balanced equation for this reaction is:
2 NaCl + H2CO3 -> Na2CO3 + 2 HCl
From the balanced equation, we can see that 2 moles of sodium chloride react to produce 1 mole of sodium carbonate.
Step 1: Calculate the molar mass of sodium chloride (NaCl) and sodium carbonate (Na2CO3):
- Molar mass of NaCl = atomic mass of Na (22.99 g/mol) + atomic mass of Cl (35.45 g/mol) = 58.44 g/mol
- Molar mass of Na2CO3 = (2 * atomic mass of Na) + atomic mass of C + (3 * atomic mass of O) = (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol) = 105.99 g/mol
Step 2: Convert the given mass of sodium chloride into moles:
- Moles of NaCl = given mass (25 g) / molar mass of NaCl (58.44 g/mol) = 0.428 mol
Step 3: Use the stoichiometric ratio to calculate the moles of sodium carbonate:
- Moles of Na2CO3 = (moles of NaCl) / (stoichiometric ratio) = 0.428 mol / 2 = 0.214 mol
Step 4: Calculate the theoretical yield of sodium carbonate in grams:
- Theoretical yield of Na2CO3 = (moles of Na2CO3) * (molar mass of Na2CO3) = 0.214 mol * 105.99 g/mol = 22.68 g
Therefore, the theoretical yield of sodium carbonate is 22.68 grams.
To calculate the percent yield, we need to compare the actual yield obtained by the student (17.9 grams) to the theoretical yield (22.68 grams).
Step 1: Calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) * 100% = (17.9 g / 22.68 g) * 100% = 78.9%
Therefore, the percent yield obtained by the student is 78.9%.