Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produc CaCO3(s) + 2 HCl(aq) CaCl (aq) + H2O(l) + CO2(g) How many grams of calcium chloride will be produced when 27.0 g of calcium carbonate is combined with 12. hydrochloric acid? mass of CaCl2: Which reactant is in excess? O Caco, OHCI How many grams of the excess reactant will remain after the reaction is complete? mass of excess reactant:

Answer 1

Answer:

Explanation:

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio can be determined by examining the coefficients in front of formulas in a balanced chemical equation.

Step 1: write the balanced chemical equation.

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

Step 2: calculate the moles of CaCO₃(s) and HCl(aq)

Moles of CaCO₃ = mass given / molar mass = ( 27 g / 100.0869 g/mol ) = 0.269766 mol

Moles of HCl = (12 g / 36.46094 g/mol ) = 0.32912 mol

Step 3: Determine the limiting reagent

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

According to the reaction we need
2 mol of HCl requires for 1 mol of CaCO₃
so, for 0.32912 mol of HClwe will need =(1 mol of CaCO₃ / 2 mol of HCl)× 0.32912 mol of HCl = 0.16456 mol of CaCO₃
Since we need only 0.16456 mol of CaCO₃ hence CaCO₃ is in excess ( since given = 0.269766 mol ) and HCl limiting reagent.

so the amount of product will formed according to the limiting reagent ( i.e HClamount )

Step 4: Calculate the moles of calcium chloride produced

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

According to the reaction:
2 mol of HCl produce 1 mol of CaCl₂
so, 0.32912 mol of HCl will produce=( 1 mol of CaCl₂ / 2 mol of HCl ) × 0.32912 mol of HCl = 0.16456 mol of CaCl₂

(a) Step 5: Calculation of mass of CaCl₂ produced

we get moles of CaCl₂ that can be produced = 0.16456 mol

Mass of CaCl₂ produced =( moles×molar mass ) = ( 0.16456 mol × 110.98 g/mol ) = 18.26 g

(c) Step 6: Calculation of excess   CaCO₃ left:

we know moles of used = 0.269766 mol -  0.16456 mol = 0.105206 mol

hence, mass remaining after reaction in excess = 0.105206 mol × 100.0869 g/mol = 10.53 g

Answer:

1. Mass of CaCl₂ produced = 18.26 g

2. Excess reagent = CaCO₃

3. Excess reagent left = 10.53 g