PART B Answers
1. How many grams of copper (I) sulfide could be produced from 9.9 grams of copper (I) chloride reacting with excess hydrosulfuric acid?
2 CuCl + H2S à Cu2S + 2 HCl
9.9 g CuCl 1 mol CuCl 1 mol Cu2S 159.16 g Cu2 S
= 7.96 g Cu2S
99 g CuCl 2 mol CuCl 1 mol Cu2S
2. 9 grams of aluminum reacts with 8 grams of sulfur to produce aluminum sulfide.
2 Al + 3 S à Al2S3
Which reactant is the limiting reactant? Pick one of the reactants and determine how much of the other reactant would be needed if all of the first reactant were used:
9 g Al 1 mol Al 3 mol S 32.06 g S
= 16.04 g S
26.98 g Al 2 mol Al 1 mol S
If all 9 grams of Al were reacted, 16.04 grams of Sulfur would be needed. Since there is only 8 grams of sulfur available, sulfur is the limiting reactant.
What mass of aluminum sulfide will be produced?
8 g S 1 mol S 1 mol Al2S3 150.14 g Al2S3
= 12.49 g Al2S3
32.06 g S 3 mol S 1 mol Al2S3
Which reactant will be in excess and by how much?
8 g S 1 mol S 2 mol Al 26.98 g Al
= 4.49 g Al
32.06 g S 3 mol S 1 mol Al
4.49 g of Al are consumed when all 8 grams of S are reacted. Since there were 9 grams of Al to start with, it is in excess by 4.51 g.
3. The compound “cisplatin” PtCl2(NH3)2, has been found to be effective in treating some types of cancer. It can be synthesized using the following equation:
K2PtCl4 + NH3 ------> KCl + PtCl2(NH3)2
How much cisplatin can be produced from 2.5 grams of K2PtCl4?
2.5 g K2PtCl4 1 mol K2PtCl4 1 mol PtCl2(NH3)2 300.06 g PtCl2(NH3)2
415.08 g 1 mol K2PtCl4 1 mol PtCl2(NH3)2
K2PtCl4
= 1.8 g PtCl2(NH3)2
How much NH3 would be needed?
2.5 g K2PtCl4 1 mol K2PtCl4 2 mol NH3 17.04 g NH3
= .21 gNH3
415.08 g 1 mol K2PtCl4 1 mol NH3
K2PtCl4
4. Cl2 + HgO ----> HgCl2 + Cl2O
2 Cl2 + HgO ----> HgCl2 + Cl2O (Balanced)
Which reactant is the limiting reactant if 116 L of Cl2 and 7.62 grams of HgO react?
116 L Cl2 1 mol Cl2 1 mol HgO 216.59 g HgO
= 560.8 g HgO
22.4 L Cl2 2 mol Cl2 1 mol HgO
If 116 L of O2 is completely reacted, then 560.8 g of HgO would be required. Since there is only 7.62 g of HgO available, HgO is the limiting reactant.
What volume of Cl2O can be produced?
7.62 g HgO 1 mol HgO 1 mol Cl2O 22.4 L Cl2O
= .79 L Cl2O
216.59 g HgO 1 mol HgO 1 mol Cl2O
Which reactant is in excess and by how much?
7.62 g HgO 1 mol HgO 2 mol Cl2 22.4 L Cl2
= 1.58 L Cl2
216.59 g HgO 1 mol HgO 1 mol Cl2
Since 116 L of Cl2 is available and only 1.58 L of Cl2 is consumed, there would be an excess of 114.42 L Cl2