Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

CaCO3 + 2HCl ⟶CaCl2 + H2O + CO2

A) How many grams of calcium chloride will be produced when 26.0g of calcium carbonate are combined whith 12.0g of hydrochloric acid?
B) Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

Answer 1

Answer: a) 18.3 grams

b)
`CaCO_3` is the excess reagent and 16.5g of
`CaCO_3` will remain after the reaction is complete.

Step-by-step explanation:

`CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2`

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

`\text{Number of moles of calcium carbonate}=(26g)/(100g/mol)=0.26moles`

`\text{Number of moles of hydrochloric acid}=(12g)/(36.5g/mol)=0.33moles`

According to stoichiometry:

2 mole of
`HCl` react with 1 mole of
`CaCO_3`

0.33 moles of
`HCl` will react with=
`(1)/(2)* 0.33=0.165moles` of
`CaCO_3`

Thus
`HCl` is the limiting reagent as it limits the formation of product and
`CaCO_3` is the excess reagent.

2 moles of
`HCl` produce = 1 mole of
`CaCl_2`

0.33 moles of
`HCl` produce=
`(1)/(2)* 0.33=0.165moles` of
`CaCl_2`

Mass of
`CaCl_2=moles* {\text{Molar Mass}}=0.165* 111=18.3g`

As 0.165 moles of
`CaCO_3` are used and (0.33-0.165)=0.165 moles of
`CaCO_3` are left unused.

Mass of
`CaCO_3` left unreacted =
`moles* {\text {Molar mass}}=0.165* 100=16.5g`

Thus 18.3 g of
`CaCl_2` are produced.
`CaCO_3` is the excess reagent and 16.5g of
`CaCO_3` will remain after the reaction is complete.