Question

Homework 3 Write the balanced equations. Calculate how many grams of each reactant will be needed to obtain 100.0 grams of the insoluble product formed in the reaction. Show complete solutions. 1. aluminum chloride + calcium hydroxide  aluminum hydroxide + calcium chloride 2. mercury (II) oxide  mercury + oxygen 3. barium nitrate + copper (II) sulfate  barium sulfate+ copper (II) nitrate 4. lead (II) chloride + potassium iodide  lead (II) iodide + potassium chloride 5. sodium sulfide + copper (II) chloride  copper (II) sulfide + sodium chloride

Answer 1

Answer:

1. 170,9 g of AlCl₃ and 142,5 g of Ca(OH)₂

2. 108,0 g of HgO

3. 112,0 g of Ba(NO₃)₂ and 68,39 g of CuSO₄

4. 60,32 g of PbCl₂ and 72,01 g of KI

5. 81,63 g of Na₂S and 140,6 g of CuCl₂

Explanation:

1. The reaction is:

2 AlCl₃ + 3 Ca(OH)₂ → 2 Al(OH)₃ + 3 CaCl₂

The insoluble product is Al(OH)₃. To produce 100,0 g you need to add:

100,0 g ×$\frac{1 mol}{78 g}$ = 1,282 moles of Al(OH)₃.

Thus, the grams of AlCl₃ and Ca(OH)₂ you need are:

1,282 moles of Al(OH)₃ ×$\frac{2 AlCl_3 moles}{2 Al(OH)_3 moles}$ = 1,282 moles of AlCl₃ × $\frac{133,34 g}{1 mol}$ = 170,9 g of AlCl₃

1,282 moles of Al(OH)₃ ×$\frac{3 Ca(OH)_2 moles}{2 Al(OH)_3 moles}$ = 1,923 moles of Ca(OH)₂ × $\frac{74,093 g}{1 mol}$ = 142,5 g of Ca(OH)₂

2. The reaction is:

2 HgO + → 2 Hg + O₂

The insoluble product is Hg. To produce 100,0 g you need to add:

100,0 g ×$\frac{1 mol}{200,59 g}$ = 0,4985 moles of Hg.

Thus, the grams of HgO you need are:

0,4985 moles of Hg ×$\frac{2 HgO moles}{2 Hg moles}$ = 0,4985 moles of HgO × $\frac{216,59 g}{1 mol}$ = 108,0 g of HgO

3. The reaction is:

Ba(NO₃)₂ + CuSO₄ →  BaSO₄ + Cu(NO₃)₂

The insoluble product is BaSO₄. To produce 100,0 g you need to add:

100,0 g ×$\frac{1 mol}{233,38 g}$ = 0,4285 moles of BaSO₄.

Thus, the grams of Ba(NO₃)₂ and CuSO₄ you need are:

0,4285 moles of BaSO₄ ×$\frac{1 Ba(NO_{3})_2 moles}{1 BaSO₄ moles}$ = 0,4285 moles of Ba(NO₃)₂ × $\frac{261,337 g}{1 mol}$ = 112,0 g of Ba(NO₃)₂

0,4285 moles of BaSO₄ ×$\frac{1 CuSO_4 moles}{1 BaSO_4 moles}$ = 0,4285 moles of CuSO₄ × $\frac{159,609 g}{1 mol}$ = 68,39 g of CuSO₄

4. The reaction is:

PbCl₂ + 2 KI →  PbI₂ + 2 KCl

The insoluble product is PbI₂. To produce 100,0 g you need to add:

100,0 g ×$\frac{1 mol}{461,01 g}$ = 0,2169 moles of PbI₂.

Thus, the grams of PbCl₂ and KI you need are:

0,2169 moles of PbI₂ ×$\frac{1 PbCl_2 moles}{1 PbI_2 moles}$ = 0,2169 moles of PbCl₂ × $\frac{278,1 g}{1 mol}$ =  60,32 g of PbCl₂

0,2169 moles of PbI₂ ×$\frac{2 KI moles}{1 PbI_2 moles}$ = 0,4338 moles of KI × $\frac{166,0028 g}{1 mol}$ = 72,01 g of KI

5. The reaction is:

Na₂S +  CuCl₂ → CuS + 2 NaCl

The insoluble product is CuS. To produce 100,0 g you need to add:

100,0 g ×$\frac{1 mol}{95,611 g}$ = 1,046 moles of CuS.

Thus, the grams of Na₂S and CuCl₂ you need are:

1,046 moles of CuS ×$\frac{1 Na_{2}S moles}{1 CuS moles}$ = 1,046 moles of Na₂S × $\frac{78,0452 g}{1 mol}$ = 81,63 g of Na₂S

1,046 moles of CuS ×$\frac{1 CuCl_2 moles}{1 CuS moles}$ = 1,046 moles of CuCl₂ × $\frac{134,45 g}{1 mol}$ = 140,6 g of CuCl₂

I hope it helps!