Calcium carbonate reacts with hydrochloric acid to form calcium chloride, carbon dioxide, and water. What mass of hydrochloric acid is needed to produce 5.00 x 10^3 kg of calcium chloride?

To solve this problem, we will use the reaction equation and stoichiometry. The balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid is:

CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)

From the balanced equation, we observe that one mole of calcium carbonate reacts with two moles of hydrochloric acid to produce one mole of calcium chloride, one mole of carbon dioxide, and one mole of water.

First, let's find the molar mass of the reactants and products:

- Calcium carbonate (CaCO3): Ca (40.08 g/mol) + C (12.01 g/mol) + 3O (3 x 16.00 g/mol) = 100.09 g/mol - Hydrochloric acid (HCl): H (1.01 g/mol) + Cl (35.45 g/mol) = 36.46 g/mol - Calcium chloride (CaCl2): Ca (40.08 g/mol) + 2Cl (2 x 35.45 g/mol) = 110.98 g/mol

Given that we want to produce 5.00 x 10^3 kg (or 5.00 x 10^6 grams, because 1 kg = 1000 g) of calcium chloride, we can find out how many moles this corresponds to:

Moles of CaCl2 = mass (g) / molar mass (g/mol) Moles of CaCl2 = 5.00 x 10^6 g / 110.98 g/mol Moles of CaCl2 ≈ 45045.95 mol

According to the stoichiometry of the reaction, 1 mole of calcium chloride is produced from 2 moles of hydrochloric acid. Hence, we need twice as many moles of HCl to react with the calcium chloride:

Moles of HCl needed = 2 x Moles of CaCl2 Moles of HCl needed = 2 x 45045.95 mol Moles of HCl needed ≈ 90091.90 mol

Now, we can find the mass of hydrochloric acid needed:

Mass of HCl = Moles of HCl x Molar mass of HCl Mass of HCl = 90091.90 mol x 36.46 g/mol Mass of HCl ≈ 3286047.03 g

Since the problem asked for the mass in kilograms, we convert grams to kilograms:

Mass of HCl = 3286047.03 g / 1000 g/kg Mass of HCl ≈ 3286.05 kg

Therefore, approximately 3286.05 kg of hydrochloric acid is needed to produce 5.00 x 10^3 kg of calcium chloride.