25.0 g of sodium carbonate is combined with an excess of hdrochloric acid. in the end 5.70 g of sodium chloride is isolated. what is the percent yield of sodium chloride
A worked example. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the percent yield of sodium chloride, you need to compare the actual yield (5.70 g) to the theoretical yield.
1. Determine the balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl):
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
2. Calculate the molar mass of sodium carbonate (Na2CO3) and sodium chloride (NaCl):
Na2CO3: 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol
NaCl: 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
3. Convert the given mass of sodium carbonate (25.0 g) to moles:
Moles of Na2CO3 = Mass / Molar mass
Moles of Na2CO3 = 25.0 g / 105.99 g/mol = 0.236 mol
4. Use stoichiometry from the balanced equation to find the theoretical yield (in moles) of sodium chloride:
From the balanced equation, the molar ratio of Na2CO3 to NaCl is 1:2.
Moles of NaCl = 2 * Moles of Na2CO3
Moles of NaCl = 2 * 0.236 mol = 0.472 mol
5. Convert the theoretical yield of sodium chloride from moles to grams:
Mass of NaCl = Moles * Molar mass
Mass of NaCl = 0.472 mol * 58.44 g/mol = 27.56 g
6. Calculate the percent yield of sodium chloride:
Percent Yield = (Actual yield / Theoretical yield) * 100%
Percent Yield = (5.70 g / 27.56 g) * 100% = 20.7%
Therefore, the percent yield of sodium chloride is approximately 20.7%.