INSTRUCTION: Give your answers in backside of the page. calculations to get cred

Question

INSTRUCTION: Give your answers in backside of the page. calculations to get credit. You MUST show your Calculate the mass (in g) of AgCI that can be formed from the reaction of 85.0 mL of 0.110 M MgCl2 solution with 65.0 mL of a 0.150 M AgCH,O2 solution. (2 Pts.) 1. 2. Calculate the molarity of a solution prepared by dissolving 2.500 g of Na2COs in water to make 500 mL of solution. (2 pts.) 3. Calculate the number of chloride ions present in 60.0 mL of 0.250 M AICI solution. (3 Pts.) 4. Considering the following reaction: Na2SO4 (s) + 4 C(s)--Na2S (s) + 4 CO (g), calculate the mass of Na2S that can be produced from a reaction mixture of 15.0 g Na2SO4 and 17.5 g C. (3 Pts.)

Explanation / Answer

1)

85.0 ml 0.110 M MgCl2 contains = (0.110*85.0)/1000

= 0.0095 moles MgCl2

65.0 ml 0.150 M AgC2H3O2 contains = (0.150*65.0)/1000

= 0.0098 moles AgC2H3O2

Stoichiometric equation between MgCl2 and AgC2H3O2 can be written as

MgCl2 + 2 AgC2H3O2 -----------> Mg(C2H3O2)2 + 2 AgCl

According to stoichiometry 1 mole MgCl2 requires 2 moles of AgC2H3O2.

But here 0.0095 mole MgCl2 and 0.0098 mole AgC2H3O2 is present.

Actually 0.0095 mole MgCl2 needs 0.019 mole AgC2H3O2 , so here AgC2H3O2 is used completely in the reaction and hence is the limiting reagent.

So here 0.0098 mole AgC2H3O2 reacts completely to form 0.0098 mole AgCl. {According to stoichiometry }

Mass in g = no.of moles * molar mass

Molar mass of AgCl = 143.32 g/mol

Mass of AgCl = 0.0098 mol * 143.32 g/mol = 1.405 g

2)

Molarity = no.of moles / volume in litre

Molar mass of Na2CO3 = 105.99 g/mol

No.of moles = mass in g/ molar mass

Mass of Na2CO3 = 2.500 g

Volume in litre = 0.500 L

No.of moles = 2.500 g/ (105.99 g/mol) = 0.0236 mol

Molarity = (0.0236 mol)/0.500 L = 0.015 M

3)

AlCl3 ionises as Al^+ and 3Cl^-.

No.of moles in 60.0 ml 0.250 M AlCl3 = (0.250*60)/1000

= 0.015 moles of AlCl3

Thus 0.015 mole AlCl3 contains 0.045 moles Cl^-.

We know 1 mole ion contains 6.02 *10^23 number of ions.

Hence 0.045 mole contains = 0.045 * 6.02 *10^23 = 2.71 * 10^22 ions

Number of chloride ions present = 2.71 * 10^22

4)

Na2SO4 (s) + 4 C (s) -----------> Na2S (s) + 4 CO (g).

15.0 g Na2SO4 = (15)/(142.04) = 0.106 moles

17.5 g C = (17.5)/(12) = 1.45 moles.

According to stoichiometry 1 mole Na2SO4 requires 4 moles C.

Hence 0.106 moles Na2SO4 requires 0.424 moles C.So here Na2SO4 is used completely in the reaction and hence is the limiting reagent.

Thus 0.106 moles Na2SO4 produces 0.106 moles Na2S.

Molar mass of Na2S = 78.05 g/mol

0.106 moles Na2S =(0.106 * 78.05) g Na2S = 8.3 g

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