Final answer:
To produce 10.0g of KCl from the reaction between calcium chloride and potassium carbonate, 7.44g of calcium chloride are needed, based on stoichiometry calculations and the molar mass of the compounds involved.
Step-by-step explanation:
The question is asking how many grams of calcium chloride (CaCl₂) are needed to produce 10.0g of potassium chloride (KCl) from the reaction CaCl₂(aq) + K₂CO₃(aq) → 2 KCl(aq) + CaCO₃(s). To determine this, we must use stoichiometry based on the balanced chemical equation. The molar mass of calcium chloride is 110.98 g/mol, and the molar mass of potassium chloride is 74.55 g/mol. Using the molar ratio from the balanced equation, which is 1 mole of CaCl₂ to 2 moles of KCl, we can calculate the mass of calcium chloride needed for 10.0 g of KCl.
To begin, we calculate the moles of KCl:
10.0 g KCl × (1 mol KCl / 74.55 g KCl) results in approximately 0.134 moles of KCl. Then we use the stoichiometric ratio from the equation to find the moles of CaCl₂ needed, which is half the moles of KCl, so it's 0.067 moles of CaCl₂. Finally, converting moles of CaCl₂ to grams:
0.067 mol CaCl₂ × (110.98 g CaCl₂ / 1 mol CaCl₂) results in about 7.44g of CaCl₂ needed.