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How many grams of Ag2CO3 will precipitate when excess Na2CO3 solution is added to 79.0 mL of 0.750 M AgNO3 solution

Question

How many grams of Ag2CO3 will precipitate when excess Na2CO3 solution is added to 79.0 mL of 0.750 M AgNO3 solution

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Step 1

The molarity of a solution is the ratio of number of moles of a substance to the volume of solution in liters. The expression for molarity is labelled in equation (1) in which M is the molarity, n is the number of moles of solute and  V is the volume in liters.

The relation between millilitres (mL) and liters (L) is written in equation(2).

The number of moles of a substance is the ratio of its given mass by molar mass of the substance. It is expressed in equation (3) in which n is the number of moles, m is the given mass and M is the molar mass.

М-.
V(L)
(1)
.. (2)
1 L 1000 mL
(3)
n =
E Z
l
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М-. V(L) (1) .. (2) 1 L 1000 mL (3) n = E Z l

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Step 2

Convert 79.0 mL in L by using the equation (2) as shown below.

1 L
79 mL 79 mL
1000 mL
=0.079 L
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1 L 79 mL 79 mL 1000 mL =0.079 L

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Step 3

The reaction of sodium carbonate and silver nitrate is written below.

2AgNO3(aq) + Na2CO3(aq) → Ag2CO3 (s) +2NaNO3(aq)

The number of moles of sil...

n
0.750 M=
0.079 L
n 0.0592 mol
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n 0.750 M= 0.079 L n 0.0592 mol

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