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 and percent yield, we need to use stoichiometry and the balanced chemical equation for the reaction between sodium chloride (NaCl) and sodium carbonate (Na2CO3).
The balanced chemical equation for the reaction is:
2 NaCl + H2CO3 -> Na2CO3 + 2 HCl
First, we need to find the molar mass of sodium chloride (NaCl) and sodium carbonate (Na2CO3). The molar mass of sodium chloride is 58.44 g/mol (22.99 g/mol for sodium + 35.45 g/mol for chlorine), and the molar mass of sodium carbonate is 105.99 g/mol (22.99 g/mol for sodium + (3 * 16.00 g/mol) for carbon + (3 * 16.00 g/mol) for oxygen).
Next, we need to calculate the number of moles of sodium chloride given the mass of 25 grams. This can be done using the formula: moles = mass / molar mass.
moles of NaCl = 25 g / 58.44 g/mol = 0.428 mol
According to the balanced chemical equation, the ratio between sodium chloride and sodium carbonate is 2:1. Therefore, the number of moles of sodium carbonate formed will be half of the moles of sodium chloride.
moles of Na2CO3 = 0.428 mol * 0.5 = 0.214 mol
Now, we can calculate the theoretical yield of sodium carbonate using the formula: theoretical yield = moles * molar mass.
theoretical yield = 0.214 mol * 105.99 g/mol = 22.68 g
Therefore, the theoretical yield of sodium carbonate is 22.68 grams.
To find the percent yield, we compare the actual yield (17.9 g) obtained by the student to the theoretical yield (22.68 g) and calculate the percentage using the formula: percent yield = (actual yield / theoretical yield) * 100.
percent yield = (17.9 g / 22.68 g) * 100 = 78.9%
So, the percent yield obtained by the student is 78.9%.