How many grams of Calcium Chloride can be formed by reacting 50.05 g of Calcium carbonate with 900.00 ml of .8000 M hydrochloric acid
To determine the grams of Calcium Chloride (CaCl2) formed from the reaction between Calcium Carbonate (CaCO3) and hydrochloric acid (HCl), we need to follow a few steps.
Step 1: Write the balanced chemical equation for the reaction:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
Step 2: Calculate the number of moles of CaCO3:
We can use the formula n = m/M, where n is the number of moles, m is the mass, and M is the molar mass.
Given that the mass of CaCO3 is 50.05 grams and its molar mass is 100.09 g/mol:
n(CaCO3) = 50.05 g / 100.09 g/mol
n(CaCO3) ≈ 0.500 mol
Step 3: Determine the limiting reactant:
To find the limiting reactant, we compare the moles of CaCO3 to HCl using the stoichiometry of the balanced equation.
From the balanced equation, we can see that 1 mole of CaCO3 reacts with 2 moles of HCl to form 1 mole of CaCl2.
So, the moles of HCl needed to react with the given amount of CaCO3 can be calculated as follows:
n(HCl) = 0.500 mol CaCO3 * (2 mol HCl / 1 mol CaCO3)
n(HCl) = 1.000 mol
However, we need to convert the given volume of HCl solution (900.00 mL) into moles. Since the concentration of the HCl solution is given as 0.8000 M (moles per liter), we have:
n(HCl) = 0.8000 mol/L * 0.90000 L
n(HCl) = 0.720 mol
From the comparison, we can see that 0.720 mol of HCl is the limiting reactant because it is lower than the moles of CaCO3.
Step 4: Calculate the moles and then grams of CaCl2 produced:
Using the stoichiometry relationship, we know that for every 2 moles of HCl reacted, 1 mole of CaCl2 is produced. Therefore, the moles of CaCl2 produced can be calculated as follows:
n(CaCl2) = 0.720 mol HCl * (1 mol CaCl2 / 2 mol HCl)
n(CaCl2) = 0.360 mol
Finally, we can calculate the mass of CaCl2 using the molar mass of CaCl2, which is 110.98 g/mol:
m(CaCl2) = n(CaCl2) * M(CaCl2)
m(CaCl2) = 0.360 mol * 110.98 g/mol
m(CaCl2) ≈ 39.87 g
So, approximately 39.87 grams of Calcium Chloride (CaCl2) can be formed by reacting 50.05 grams of Calcium Carbonate (CaCO3) with 900.00 mL of 0.8000 M hydrochloric acid (HCl).