# molar mass

A certain sample of calcium carbonate(CaCO3) contains 4.86mol.

a) What is the mass in grams of this sample?

b) What is the mass of the CO3^2- ions present?

calculate the mass of magnesium oxide (MgO) formed when 2.43g of maganesium (Mg) are burnt with

a) excess ocygen

b)1.28g of oxygen

A mixture containing 2.8g of iron and 2g sulphur is heated together.

What is the mass of iron(II) sulphide, FeS, is produced?

Fe(s) + S(s) -->FeS(s)

ans 係咪. 4.4048g?

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1.

A certain sample of calcium carbonate(CaCO3) contains 4.86 mol.

a) What is the mass in grams of this sample?

b) What is the mass of the CO3²⁻ ions present?

a)

Molar mass of CaCO3 = 40 + 12 + 16x3 = 100 g/mol

Mass of CaCO3­ = (4.86 mol) x (100 g/mol) = 486 g

b)

No. of moles of CO3²⁻ ions = No. of moles of CaCO3 = 4.86 mol

Molar mass of CO3²⁻ ions = 12 + 16x3 = 60 g/mol

Molar mass of CO3²⁻ ions = (4.86 mol) x (60 g/mol) = 291.6 g

= = = = =

2.

calculate the mass of magnesium oxide (MgO) formed when 2.43g of maganesium (Mg) are burnt with

a) excess oxygen

b)1.28g of oxygen

a)

Molar mass of Mg = 24.3 g/mol

Molar mass of O2 = 16x2 = 32 g/mol

Molar mass of MgO = 24.3 + 16 = 40.3 g/mol

2Mg + O2 → 2MgO

Molar Mass Mg : MgO = 2 : 2 = 1 : 1

No. of moles of Mg = (2.43 g) / (24.3 g/mol) = 0.1 mol

No. of moles of MgO formed = 0.1 mol

Mass of MgO formed = (0.1 mol) x (40.3 g/mol) = 4.03 g

b)

2Mg + O2 → 2MgO

Molar Mass Mg : O2 : MgO = 2 : 1 : 2

No. of moles of Mg present = 0.1 mol

No. of moles of O2 present = (1.28 g)/ (32 g/mol) = 0.04 mol

To complete react with 0.04 mol of O2,

no. of moles of Mg needed = (0.04 mol) x 2 = 0.08 mol

Hence, Mg is in excess, and O2 completely reacts (limiting reagent).

No. of moles of O2 reacted = 0.04 mol

No. of moles of MgO formed =- (0.04 mol) x 2 = 0.08 mol

Mass of MgO formed = (0.08 mol) x (40.3 g/mol) = 3.224 g

= = = = =

3.

A mixture containing 2.8 g of iron and 2 g sulphur is heated together. What is the mass of iron(II) sulphide, FeS, is produced?

Fe(s) + S(s) → FeS(s)

Molar mass of Fe = 55.8 g/mol

Molar mass of S = 32.1 g/mol

Molar mass of FeS = 55.8 + 32.1 = 87.9 g/mol

Mole ratio Fe : S : FeS = 1 : 1 : 1

No. of moles of Fe present = 2.8/55.8 = 0.0502 mol

No. of moles of S present = 2/32.1 = 0.0623 mol

Hence, Fe is the limiting reagent (completely reacted).

No. of moles of Fe reacted = 0.0502 mol

No. of moles of FeS formed = 0.0502 mol

Mass of FeS formed = 0.0502 x 87.9 = 4.41 g

資料來源： micatkie