When solid lead(ii) sulfide ore burns in oxygen gas, the products are solid lead(ii) oxide and sulfur dioxide gas.? a. write the balanced equation b. how many grams of oxygen are required to react with 29.9 grams of lead ii sulfide? c. how many grams of sulfur oxide can be produced when 65.0 grams of lead ii sulfide reacts? d. how many grams of lead ii sulfide are used to produce 128 grams of lead ii oxide?

A. The complete balanced chemical reaction is:

PbS  +  1.5 O2  --->  PbO  +  SO2

B. First let us convert mass of PbS into number of moles. The molar mass of PbS is 239.3 g/mol, hence:

moles PbS = 29.9 g/ (239.3 g/mol) = 0.125 mol

From the reaction, we need 1.5 moles of O2 for every 1 mole of PbS, therefore:

moles O2 = 0.125 mol * 1.5 = 0.1875 mol

The molar mass of O2 is 32 g/mol, hence the mass is:

mass O2 = 0.1875 mol * 32 g/mol = 6 grams O2

C. Converting mass to number of moles:

moles PbS = 65.0 g/ (239.3 g/mol) = 0.2716 mol

From the reaction, we can produce 1 mole of SO2 for every 1 mole of PbS, therefore:

moles SO2 = 0.2716 mol

The molar mass of O2 is 64 g/mol, hence the mass is:

mass SO2 = 0.2716 mol * 64 g/mol = 17.38 grams SO2

D. First let us convert mass of PbO into number of moles. The molar mass of PbO is 223.2 g/mol, hence:

moles PbO = 128 g/ (223.2 g/mol) = 0.573 mol

From the reaction, we need 1 mole of PbS for every 1 mole of PbO, therefore:

moles PbS = 0.573 mol

The molar mass of PbS is 239.3 g/mol, hence the mass is:

mass PbS = 0.573 mol * 239.3 g/mol = 137.23 grams PbS

c

Explanation: