Ex 2: Aluminium chloride reacts with tin(IV)oxide to producealuminium oxide and tin(IV)chloride.a) What mass of aluminium chloride is required to react with1.00 g of the tin oxide?b) What mass of each product would you expect from thereaction in (a)?Follow Up Problems3.8Problems3.61, 63, 65, 67

To solve this problem, we first need to write down the balanced chemical equation for the reaction:

2AlCl3 + SnO2 → 2Al2O3 + SnCl4

a) To find the mass of aluminium chloride required to react with 1.00 g of tin(IV) oxide, we first need to find the molar mass of SnO2 and AlCl3.

The molar mass of SnO2 is approximately 150.71 g/mol (118.71 g/mol for Sn and 32.00 g/mol for O2).

The molar mass of AlCl3 is approximately 133.34 g/mol (26.98 g/mol for Al and 35.45 g/mol for Cl3).

Next, we calculate the number of moles of SnO2 in 1.00 g:

1.00 g SnO2 * (1 mol SnO2 / 150.71 g SnO2) = 0.00664 mol SnO2

From the balanced equation, we know that 2 moles of AlCl3 react with 1 mole of SnO2. Therefore, we need:

0.00664 mol SnO2 * (2 mol AlCl3 / 1 mol SnO2) = 0.0133 mol AlCl3

Finally, we convert this amount to grams:

0.0133 mol AlCl3 * (133.34 g AlCl3 / 1 mol AlCl3) = 1.77 g AlCl3

So, 1.77 g of aluminium chloride is required to react with 1.00 g of tin(IV) oxide.

b) To find the mass of each product, we use the stoichiometry of the balanced equation. For every 2 moles of AlCl3 that react, we get 2 moles of Al2O3 and 1 mole of SnCl4.

The molar mass of Al2O3 is approximately 101.96 g/mol (26.98 g/mol for Al and 16.00 g/mol for O3).

The molar mass of SnCl4 is approximately 260.52 g/mol (118.71 g/mol for Sn and 35.45 g/mol for Cl4).

So, the mass of Al2O3 produced is:

0.0133 mol AlCl3 * (1 mol Al2O3 / 1 mol AlCl3) * (101.96 g Al2O3 / 1 mol Al2O3) = 1.36 g Al2O3

And the mass of SnCl4 produced is:

0.0133 mol AlCl3 * (1 mol SnCl4 / 2 mol AlCl3) * (260.52 g SnCl4 / 1 mol SnCl4) = 1.73 g SnCl4

So, we would expect 1.36 g of aluminium oxide and 1.73 g of tin(IV) chloride from the reaction.