1) 56.0 mL of 3.80 M sodium hydroxide is combined with 42.0 mL of 1.00 M magnesi

Question

1) 56.0 mL of 3.80 M sodium hydroxide is combined with 42.0 mL of 1.00 M magnesium chloride. What mass in grams of solid forms?

2) Sodium thiosulfate, Na2S2O3, is used as a fixer in photographic film developing. The amount of Na2S2O3 in a solution can be determined by a titration with iodine, I2, according to the equation: 2Na2S2O3(aq) + I2(aq) --> Na2S4O6 +2NaI(aq).

Calculate the concentration of the Na2S2O3 solution if 33.10 mL of a 0.2590 M I2 solution react exactly with a 100.0 mL sample of the Na2S2O3 solution. Use 4 sig. fig.

3) A 0.3120 g sample of a compound made up of aluminum and chlorine is dissolved in 150 mL H2O. When excess AgNO3 is added, 1.006 g of AgCl is precipitated. What is the number of moles Al in the aluminum chloride sample?

Explanation / Answer

1- 2NaOH + MgCl2 ----> 2NaCl + Mg(OH)2

So, 1 mole of MgCl2 reacts with 2 moles of NaOH, to give 1 mole of Mg(OH)2 and 2 moles of NaCl.

Now,

millimoles of NaOH = 56*3.8 =212.8

millimoles of MgCl2 =42*1 = 42

From above balanced chemical equation, MgCl2 is the limiting reagent,

so, expected millimoles of NaCl to be formed = 42*2 = 84

expected millimoles of Mg(OH)2 = 42

Mass of NaCl to be formed = 84*58.44/1000 = 4.908 g

Mass of Mg(OH)2 to be formed = 42*58.319/1000 = 2.449 g

2- 2Na2S2O3(aq) + I2(aq) --> Na2S4O6 +2NaI(aq)

From the balanced chemical equation, 2 moles of Na2S2O3 reacts with 1 moles of I2 for complete reaction.

millimoles of I2 = 33.1*0.259 = 8.5729

Thus, for complete reaction expected millimoles of Na2S2O3 equired = 8.5729*2 = 17.1458

Thus, desired concentration of Na2S2O3 = 17.1458/ Volume (in ml) = 17.1458/100 =0.171458 M

3- AlCl3 + 3AgNO3 ----> 3AgCl + Al(NO3)3

Millimoles of AgCl formed =1.006/143.32 =7.019

expected millimoles of limiting reagent AlCl3 = 7.019/3 =2.339

No of millimoles of AlCl3 in 0.312 g of AlCl3 =0.312/133.34 =2.3398

Thus, as 1 mole of AlCl3 gived 1 mole of Al exactly.

No of moles of Al in AlCl3 sample= (2.3398/1000) = 0.0023398

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