Answer:
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
Answer:
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.
