Question

To solve stoichiometry problems, you must always calculate numbers of moles. Recall that molarity, M, is equal to the concentration in moles per liter: M=mol/L.

When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation

2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq)

Part A

What mass of silver chloride can be produced from 1.55 L of a 0.183 M solution of silver nitrate?

Express your answer with the appropriate units.

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Part B

The reaction described in Part A required 3.24 L of calcium chloride. What is the concentration of this calcium chloride solution?

Express your answer with the appropriate units.

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Answer 1

A)

volume = 1.55 L

we have below equation to be used:

number of mol ,

n = Molarity * Volume

= 0.183*1.55

= 0.2837 mol

From balanced chemical reaction, we see that

when 2 mol of AgNO3 reacts, 2 mol of AgCl is formed

mol of AgCl formed = (2/2)* moles of AgNO3

= (2/2)*0.2837

= 0.2837 mol

Molar mass of AgCl = 1*MM(Ag) + 1*MM(Cl)

= 1*107.9 + 1*35.45

= 143.35 g/mol

we have below equation to be used:

mass of AgCl = number of mol * molar mass

= 0.2837*1.434*10^2

= 40.7 g

Answer: 40.7 g

B)

we have the Balanced chemical equation as:

2 AgNO3 + CaCl2 ---> AgCl + 2 Ca(NO3)2

Here:

M(AgNO3)=0.183 M

V(AgNO3)=1.55 L

V(CaCl2)=3.24 L

According to balanced reaction:

1*number of mol of AgNO3 =2*number of mol of CaCl2

1*M(AgNO3)*V(AgNO3) =2*M(CaCl2)*V(CaCl2)

1*0.183*1.55 = 2*M(CaCl2)*3.24

M(CaCl2) = 0.0438 M

Answer: 0.0438 M