20 g of calcium chloride reacts with 20 g of sodium carbonate to form calcium carbonate and sodium chloride.
CaCl2+ Na2CO3 → CaCO3 + 2NaCl (Molar mass: Ca 40g/mol; O 16 g/mol; C 12 g/mol; Cl 35.5 g/mol)
1. Calculate the number of moles of each compound:
- Calcium chloride (CaCl2): 20 g / 110 g/mol = 0.182 moles
- Sodium carbonate (Na2CO3): 20 g / 106 g/mol = 0.189 moles
2. Determine the limiting reactant:
- To determine the limiting reactant, we need to compare the number of moles of each compound and their coefficients in the balanced equation. In this case, we have one mole of CaCl2 reacting with one mole of Na2CO3. Since the number of moles of Na2CO3 is slightly higher than that of CaCl2, it is the limiting reactant.
- Theoretical yield of CaCO3: 0.189 moles x 1 mol CaCO3/1 mol Na2CO3 x 100 g/mol = 18.9 g
3. Calculate the actual yield of CaCO3:
- We don't have information about the actual yield, so we cannot compare it with the theoretical yield.
4. Calculate the percent yield:
- % yield = (actual yield / theoretical yield) x 100%
- Since we don't have the actual yield, we cannot calculate the percent yield.
To calculate the limiting reactant, we need to compare the number of moles of each reactant.
First, let's calculate the number of moles of calcium chloride (CaCl2):
Molar mass of CaCl2 = (1 x 40 g/mol) + (2 x 35.5 g/mol) = 111 g/mol
Number of moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2
Number of moles of CaCl2 = 20 g / 111 g/mol = 0.180 moles
Now, let's calculate the number of moles of sodium carbonate (Na2CO3):
Molar mass of Na2CO3 = (2 x 23 g/mol) + 12 g/mol + (3 x 16 g/mol) = 106 g/mol
Number of moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3
Number of moles of Na2CO3 = 20 g / 106 g/mol = 0.189 moles
Now, let's compare the number of moles of each reactant. The reactant with the smallest number of moles is the limiting reactant.
In this case, calcium chloride (CaCl2) has fewer moles (0.180 moles) compared to sodium carbonate (Na2CO3) (0.189 moles). Therefore, calcium chloride is the limiting reactant.
Since the balanced equation shows that the ratio between CaCl2 and CaCO3 is 1:1, we can directly state that the number of moles formed of CaCO3 will also be 0.180 moles.
Now, to calculate the mass of calcium carbonate (CaCO3) formed, we can use the following formula:
Mass of CaCO3 = Number of moles of CaCO3 x Molar mass of CaCO3
Mass of CaCO3 = 0.180 moles x (1 x 40 g/mol + 12 g/mol + 3 x 16 g/mol)
Mass of CaCO3 = 0.180 moles x 100 g/mol
Mass of CaCO3 = 18 g
Therefore, 18 g of calcium carbonate will be formed when 20 g of calcium chloride and 20 g of sodium carbonate react.