Question

3Mg +2AlCl3 -------> 2Al +3MgCl2
a)calculate the mass of aluminium chloride that will react with 40g of magnesium.

b) calculate the mass of each product formed from the reaction of 40g of mg

2)when propane gas C3H8 is burned with oxygen 50g of water is collected.what mass of propane was burned?

3)metallic iron can be produced by reacting iron(iii) oxide with carbon as shown below.when 100g of iron(iii)oxide is reacted, 45g of iron metal is obtained.what is the percent yield of the reaction?

Answer 1

a) Mass of Mg = 40 g

Molar mass of Mg =24.30 g/mol

moles of Mg = mass / molar mass = 40 g/ ( 24.30g/mol) = 1.646 mol

From given reaction, 3 mol of Mg reacts with 2 mol of AlCl₃

So, 1.646 mol of Mg reacts with 2*1.646/3 = 1.097 mol of AlCl₃

Molar mass of AlCl₃ = 133.34 g/mol

Moles mass of AlCl₃ = moles * Molar mass = 1.097 mol*133.34 g/mol = 146.33 g of AlCl₃

(b) From reaction, 3 mol of Mg produces 2 mol of Al and 3 mol of MgCl₂

So,  1.646 mol of Mg  produces 2*1.646/3 =1.097 mol of Al and 1.646 mol of MgCl₂

Molar mass of Al = 26.98 g/mol

Mass of Al produced = moles* Molar mass = 1.097 mol * 26.98 g/mol = 29.60 g

Molar mass of MgCl₂ = 95.211 g/mol

Mass of MgCl₂ produced = 1.646 mol *95.211 g/mol = 156.72 g

2. Reaction of combustion of propane is

C₃H₈ (g) + 5O₂(g) 3CO₂ (g) + 4H₂O(g)

Mass of water = 50 g

Molar mass of water = 18.02 g/mol

Moles of water = 50g/(18.02g/mol) = 2.77 mol

From reaction 4 mol of water is produced from 1 mol of C₃H₈

Thus, 2.77 mol of water is produced from 0.694  mol of C₃H₈

Molar mass of C₃H₈ = 44.1 g/mol

Mass of propane burned = moles * molar mass = 0.694 mol*44.1 g/mol = 30.59 g

3. Reaction is:

Fe₂O₃ + 3 C --> 2 Fe + 3 CO

Mass of Fe₂O₃ = 100 g

Molar mass of Fe₂O₃ = 159.69 g/mol

Moles of Fe₂O₃ = mass/ Molar mass = 100g/(159.69g/mol) = 0.626 mol

From reaction , 1mol of Fe₂O₃ produces 2 mol of Fe

Thus, 0.626 mol of Fe₂O₃ produces 1.252 mol of Fe

Molar mass of Fe = 55.845 g/mol

Mass of Fe = 1.252 mol* 55.845 g/mol = 69.94 g

Mass of iron actually obtained = 45 g

percent yield = (mass of Fe / Mass of iron actually obtained)*100%

= (45 g/ 69.94 g)*100% = 64.34 %