To solve stoichiometry problems, you must always calculate numbers of moles. Rec

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.

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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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.

Hints

mass of AgCl =

SubmitMy AnswersGive Up

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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SubmitMy AnswersGive Up

Explanation / Answer

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

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

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