How many grams of calcium chloride will be produced when 28.0 g 28.0 g of calcium carbonate is combined with 14.0 g 14.0 g of hydrochloric acid?

There will be 21.31 grams of CaCl2 produced.

Explanation:

Step 1: Data given

Mass of CaCO3 = 28.0 grams

Mass of HCl = 14.0 grams

Molar mass of CaCO3 = 100.09 g/mol

Molar mass of HCl = 36.46 g/mol

Step 2: The balanced equation

CaCO3 + 2HCl → CaCl2 + H2O + CO2

Step 3: Calculate Moles of CaCO3

Moles CaCO3 = mass CaCO3 / molar mass CaCO3

Moles CaCO3 = 28.0 grams / 100.09 g/mol

Moles CaCO3 = 2.80 moles

Step 4: Calculate moles HCl

Moles HCl = 14.0 grams / 36.46 g/mol

Moles HCl = 0.384 moles

Step 5: Calculate the limiting reactant.

For 1 mol CaCO3 we need 2 moles HCl to produce 1 mol caCl2, 1 mol CO2 and 1 mol H2O

HCl is the limiting reactant. It will completely be consumed (0.384 moles)

CaCO3 is in excess. There will react 0.384/2 = 0.192 moles

There will remain 2.80 - 0.192 = 2.608 moles

Step 6: Calculate moles CaCl2

For 1 mol CaCO3 we need 2 moles HCl to produce 1 mol caCl2, 1 mol CO2 and 1 mol H2O

For 0.384 moles HCl we'll have 0.384/2 = 0.192 moles CaCl2

Step 7: Calculate mass CaCl2

Mass CaCl2 = moles CaCl2 * molar mass

Mass CaCl2 = 0.192 moles * 110.98 g/mol

Mass CaCl2 = 21.31 grams

There will be 21.31 grams of CaCl2 produced.

21.3 g of CaCl₂ are produced in this reaction

Explanation:

Reaction is this:

CaCO₃ + 2HCl → CaCl₂ + H₂CO₃

Molar mass salt : 100.08 g/m

Molar mass acid: 36.45 g/m

We have 28 g / 100.08 g/m = 0.279 moles of CaCO₃

We have 14 g / 36.45 g/m = 0.384 moles of HCl

Ratio is 1:2, so 0.279 moles of salt will need the double of moles of acid.

0.279. 2 = 0.558 moles of acid needed. (I have only 0.384 moles, so the acid is the limiting reactant)

0.384 moles of HCl produce the half of moles of CaCl₂ (This ratio is 1:2) - -> 0.192 moles

Molar mass of CaCl₂ = 110.98 g/m

Mol. molar mass = mass → 0.192 m. 110.98 g/m = 21.3 g