Ca(OH)2 reacts with CO2 to form CaCO3 and H2O.

A. How many grams of CaCO3 will be formed if 4 grams of Ca(OH)2 are used?

B. If you have 2.5 grams of Ca(OH)2 and 0.5 of CO2, which is the limiting reactant?

A=94.6g

B=Ca(OH)2

To answer these questions, we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction. Here are the steps to solve both questions:

A. To determine the mass of CaCO3 formed, we need to start with the amount of Ca(OH)2 and use the stoichiometric coefficients from the balanced equation. Here's how:

1. Write the balanced chemical equation for the reaction:
Ca(OH)2 + CO2 -> CaCO3 + H2O

2. Find the molar mass of Ca(OH)2:
Ca(OH)2: Ca (40.08 g/mol) + 2 * O (16.00 g/mol) + 2 * H (1.01 g/mol)
Molar mass = 40.08 + 2 * 16.00 + 2 * 1.01 = 74.10 g/mol

3. Use the molar mass to convert grams of Ca(OH)2 to moles:
4 g Ca(OH)2 * (1 mol Ca(OH)2 / 74.10 g Ca(OH)2) = 0.054 mol Ca(OH)2

4. Use the stoichiometric coefficients to determine the moles of CaCO3 formed:
From the balanced equation, we can see that one mole of Ca(OH)2 reacts to form one mole of CaCO3.
Therefore, 0.054 mol Ca(OH)2 will produce 0.054 mol CaCO3.

5. Convert moles of CaCO3 to grams:
Molar mass of CaCO3: Ca (40.08 g/mol) + C (12.01 g/mol) + 3 * O (16.00 g/mol)
Molar mass = 40.08 + 12.01 + 3 * 16.00 = 100.09 g/mol

0.054 mol CaCO3 * (100.09 g CaCO3 / 1 mol CaCO3) = 5.41 g CaCO3

Therefore, if 4 grams of Ca(OH)2 are used, 5.41 grams of CaCO3 will be formed.

B. To determine the limiting reactant, we compare the moles of each reactant used and calculate the theoretical yield for CaCO3 for both reactants. Here's how:

1. Use the molar mass to convert grams of Ca(OH)2 to moles:
2.5 g Ca(OH)2 * (1 mol Ca(OH)2 / 74.10 g Ca(OH)2) = 0.034 mol Ca(OH)2

2. Use the molar mass to convert grams of CO2 to moles:
0.5 g CO2 * (1 mol CO2 / 44.01 g CO2) = 0.011 mol CO2

3. Calculate the theoretical yield of CaCO3 for each reactant based on the stoichiometry:
From the balanced equation, we can see that one mole of Ca(OH)2 reacts to form one mole of CaCO3.
Therefore, both 0.034 mol Ca(OH)2 and 0.034 mol CO2 will produce 0.034 mol CaCO3.

4. Convert the theoretical yield of CaCO3 to grams:
0.034 mol CaCO3 * (100.09 g CaCO3 / 1 mol CaCO3) = 3.41 g CaCO3 (for both reactants)

Since the theoretical yield of CaCO3 is the same for both reactants (3.41 g), neither reactant is limiting.