Question

# 1) How many grams of Ag2CO3 will precipitate when excess K2CO3 solution is added to 53.0...

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

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

Only 1 question at a time please

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