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

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ =

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution =

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

kf is constant for water and its value is 1.86

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point