Question

1) How many grams of Ag₂CO₃ will precipitate when excess K₂CO₃ solution is added to 53.0 mL of 0.713 M AgNO₃ solution?

2AgNO₃(aq) + K₂CO₃(aq) = Ag₂CO₃(s) + 2KNO₃(aq)

?g

2)How many mL of 0.695 M HBr are needed to dissolve 8.81 g of MgCO₃?

2HBr(aq) + MgCO₃(s) = MgBr₂(aq) + H₂O(l) + CO₂(g)

? mL

Answer 1

1)

Balanced chemical equation is:

2AgNO3(aq) + K2CO3(aq) = Ag2CO3(s) + 2KNO3(aq)

lets calculate the mol of AgNO3

volume , V = 53 mL

= 5.3*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.713*5.3*10^-2

= 3.779*10^-2 mol

According to balanced equation

mol of Ag2CO3 formed = (1/2)* moles of AgNO3

= (1/2)*3.779*10^-2

= 1.889*10^-2 mol

This is number of moles of Ag2CO3

Molar mass of Ag2CO3,

MM = 2*MM(Ag) + 1*MM(C) + 3*MM(O)

= 2*107.9 + 1*12.01 + 3*16.0

= 275.81 g/mol

use:

mass of Ag2CO3,

m = number of mol * molar mass

= 1.889*10^-2 mol * 2.758*10^2 g/mol

= 5.211 g

Answer: 5.21 g

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