m/ibisams/mod/ibis view.php?id-4261482 stion 6 of 13 Ma eneral Chemistry 4th Edt

Question

m/ibisams/mod/ibis view.php?id-4261482 stion 6 of 13 Ma eneral Chemistry 4th Edtion University Science Books presented by Sapling Leaming When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. How many grams of calcium chloride will be produced when 290 g of calcium carbonate are combined with 14.0 g of hydrochloric acid? Number g CaCI, Which reactant is in excess and how many grams of this reactant will remain ater the reaction is complete? Number caco, O HCI R of O Pre vous Check Answer .0 Next Era

Explanation / Answer

a)

Molar mass of CaCO3 = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass of CaCO3 = 29.0 g

we have below equation to be used:

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(29.0 g)/(100.09 g/mol)

= 0.2897 mol

Molar mass of HCl = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = 14.0 g

we have below equation to be used:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(14.0 g)/(36.458 g/mol)

= 0.384 mol

1 mol of CaCO3 reacts with 2 mol of HCl

for 0.2897 mol of CaCO3, 0.5795 mol of HCl is required

But we have 0.384 mol of HCl

so, HCl is limiting reagent

we will use HCl in further calculation

Molar mass of CaCl2 = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

From balanced chemical reaction, we see that

when 2 mol of HCl reacts, 1 mol of CaCl2 is formed

mol of CaCl2 formed = (1/2)* moles of HCl

= (1/2)*0.384

= 0.192 mol

we have below equation to be used:

mass of CaCl2 = number of mol * molar mass

= 0.192*1.11*10^2

= 21.3 g

Answer: 21.3 g

b)

From balanced chemical reaction, we see that

when 2 mol of HCl reacts, 1 mol of CaCO3 is formed

mol of CaCO3 reacted = (1/2)* moles of HCl

= (1/2)*0.384

= 0.192 mol

mol of CaCO3 remaining = mol initially present - mol reacted

mol of CaCO3 remaining = 0.2897 - 0.192

mol of CaCO3 remaining = 0.0977 mol

Molar mass of CaCO3 = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

we have below equation to be used:

mass of CaCO3,

m = number of mol * molar mass

= 9.774*10^-2 mol * 100.09 g/mol

= 9.78 g

Answer: 9.78 g of CaCO3

Feel free to comment below if you have any doubts or if this answer do not work

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