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100 ml of 0.2 mol/l sodium carbonate solution and 200 ml of 0.1 mol/l calcium nitrate solution are mixed together. Calculate the mass of calcium carbonate that would precipitate and the concentration of the sodium nitrate solution that will be produced.
Accepted Answer
Step 1: Calculate the moles of sodium carbonate and calcium nitrate
Moles of sodium carbonate = 0.2 mol/L x 0.1 L = 0.02 mol
Moles of calcium nitrate = 0.1 mol/L x 0.2 L = 0.02 mol
Step 2: Determine the limiting reactant
The balanced chemical equation for the reaction is:
Na2CO3(aq) + Ca(NO3)2(aq) -→ CaCO3(s) + 2NaNO3(aq)
From the stoichiometry of the balanced equation, we can see that 1 mole of sodium carbonate reacts with 1 mole of calcium nitrate.
Since we have 0.02 mol of each reactant, they are in a 1:1 mole ratio, so neither reactant is limiting.
Step 3: Calculate the moles of calcium carbonate that will precipitate
From the balanced equation, we can see that 1 mole of sodium carbonate reacts with 1 mole of calcium carbonate.
Therefore, 0.02 mol of sodium carbonate will react with 0.02 mol of calcium carbonate.
Moles of calcium carbonate that will precipitate = 0.02 mol
Step 4: Calculate the mass of calcium carbonate that will precipitate
Mass of calcium carbonate = 0.02 mol x 100.09 g/mol = 2.00 g
Step 5: Calculate the concentration of the sodium nitrate solution that will be produced
Moles of sodium nitrate produced = 2 x 0.02 mol = 0.04 mol
Volume of the final solution = 100 mL + 200 mL = 300 mL = 0.3 L
Concentration of the sodium nitrate solution = 0.04 mol / 0.3 L = 0.133 mol/L**
Therefore, the mass of calcium carbonate that would precipitate is 2.00 g, and the concentration of the sodium nitrate solution that will be produced is 0.133 mol/L.
Moles of sodium carbonate = 0.2 mol/L x 0.1 L = 0.02 mol
Moles of calcium nitrate = 0.1 mol/L x 0.2 L = 0.02 mol
Step 2: Determine the limiting reactant
The balanced chemical equation for the reaction is:
Na2CO3(aq) + Ca(NO3)2(aq) -→ CaCO3(s) + 2NaNO3(aq)
From the stoichiometry of the balanced equation, we can see that 1 mole of sodium carbonate reacts with 1 mole of calcium nitrate.
Since we have 0.02 mol of each reactant, they are in a 1:1 mole ratio, so neither reactant is limiting.
Step 3: Calculate the moles of calcium carbonate that will precipitate
From the balanced equation, we can see that 1 mole of sodium carbonate reacts with 1 mole of calcium carbonate.
Therefore, 0.02 mol of sodium carbonate will react with 0.02 mol of calcium carbonate.
Moles of calcium carbonate that will precipitate = 0.02 mol
Step 4: Calculate the mass of calcium carbonate that will precipitate
Mass of calcium carbonate = 0.02 mol x 100.09 g/mol = 2.00 g
Step 5: Calculate the concentration of the sodium nitrate solution that will be produced
Moles of sodium nitrate produced = 2 x 0.02 mol = 0.04 mol
Volume of the final solution = 100 mL + 200 mL = 300 mL = 0.3 L
Concentration of the sodium nitrate solution = 0.04 mol / 0.3 L = 0.133 mol/L**
Therefore, the mass of calcium carbonate that would precipitate is 2.00 g, and the concentration of the sodium nitrate solution that will be produced is 0.133 mol/L.
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