Question

1. When solid lead(II) sulfide ore burns in oxygen gas, the products are solid lead(II) oxide and sulfur dioxide gas

2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)

A.How many grams of oxygen are required to react with 26.0 g of lead(II) sulfide?

B.How many grams of sulfur dioxide can be produced when 61.0 g of lead(II) sulfide reacts?

C.How many grams of lead(II) sulfide are used to produce 121 g of lead(II) oxide?

2. Pentane gas, C5H12, undergoes combustion with oxygen to produce carbon dioxide and water.

A. Write the balance equation with phases

B.How many grams of C5H12 are needed to produce 67 g of water?

C.How many grams of CO2 are produced from 26.0 g of O2?

3. 4Al(s)+3O2(g)→2Al2O3(s) A. How many moles of O2 are needed to react with 2.50 moles of Al? B. How many grams of Al2O3 are produced when 51.8 g of Al reacts? C.When Al is reacted in a closed container with 7.46 g of O2, how many grams of Al2O3 can form?

Answer 1

2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)

A 26.0 g of lead(II) sulfide (Molar mass: 239.3 g/mol) = 26/239.3 = 0.10865 Mole

0.10865 Mole of lead(II) sulfide will need = 0.10865 x 3 x 32/2 = 5.215 gm of O2 Need

B. 61.0 g of lead(II) sulfide = 61 / 239.3 = 0.25491 Mole

0.25491 Mole of lead(II) sulfide will produce = 0.25491 x 64.04 = 16.329 gm SO2

C. 121 g of lead(II) oxide (Molar mass 223.20 g/mol) = 121 / 223.2 = 0.54211 Mole

0.54211 Mole of lead(II) oxide = 0.54211 x 239.3 = 129.72 gm of PbS Need

C₅H₁₂ + 8 O₂ = 5 CO₂ + 6 H₂O

67 g of water = 67/18 = 3.72 Mole

3.72 Mole of water = 3.72 x 72.14 /6 = 44.75 gm of C₅H₁₂ need

26.0 g of O2 = 26/ 32 = 0.8125 Mole

0.8125 Mole og oxygen = 0.8125 x 5 x 44/8 = 22.34 gm CO2 produced

3. 4Al(s)+3O2(g)→2Al2O3(s)

2.50 moles of Al need 2.5 x 3/4 = 1.875 Mole of O2 need

51.8 g of Al (Atomic mass: 26.98) = 51.8 / 26.98 = 1.92 Mole

1.92 Mole of Al will produce = 1.92 x 101.96 = 195.7 gm of Al2O3 (Molar mass: 101.96) will produce.

7.46 g of O2 = 7.46 / 32 = 0.2331 Mole

0.2331 Mole of oxygen will produce = 0.2331 x 2 x 101.96 / 3 = 15.84 gm of Al2O3 can form.