Question

2. Iron reacts with oxygen gas according to the following equation:

Answer 1

Answer:

2. a) 2.67 mol.

  b) 1.33 mol.

3. 4.35 g.

4. 8.67 g.

5. a) 143.86 L.

   b) 251.75 L.

Explanation:

2. Iron reacts with oxygen gas according to the following equation:

4Fe + 3O₂ → 2Fe₂O₃ , If 2 moles of oxygen gas is used in the reaction,

• Fe reacts with O₂ according to the balanced equation:

4Fe + 3O₂ → 2Fe₂O₃,

It is clear that 4 mole of Fe react with 3 mole of O₂ to  produce 2 moles of Fe₂O₃.

a) how many moles of iron, Fe, will be required?

using cross multiplication:

3 mol of O₂ require  → 4 mol of Fe, from the stichiometry.

2 mol of O₂ require → ??? mol of Fe.

∴ The no. of moles of of Fe are required = (2 mol)(4mol)/(3 mol) = 2.67 mol.

(b) how many moles of iron(III) oxide, Fe₂O₃, will be produced?

using cross multiplication:

3 mol of O₂ produce → 2 mol of Fe₂O₃, from the stichiometry.

2 mol of O₂ require → ??? mol of Fe₂O₃.

∴ The no. of moles of of Fe₂O₃ are produced = (2 mol)(2 mol)/(3 mol) = 1.33 mol.

3. Potassium sulfate can be prepared by the reaction between dilute sulfuric acid  and potassium carbonate.

H₂SO₄ + K₂CO₃ → K₂SO₄ + CO₂ + H₂O,

Calculate the mass of potassium sulfate that can be prepared from 3.45 g of  potassium carbonate.

• H₂SO₄ reacts with K₂CO₃ according to the balanced equation:

H₂SO₄ + K₂CO₃ → K₂SO₄ + CO₂ + H₂O,

It is clear that 1 mole of H₂SO₄ reacts with 1 mole of K₂CO₃ to  produce 1 mole of K₂SO₄, 1 mole of CO₂ and 1 mole of H₂O.

Firstly, we need to calculate the no. of moles of 3.45 g of K₂CO₃:

no. of moles of K₂CO₃ = mass/molar mass = (3.45 g)/(138.205 g/mol) = 0.025 mol.

using cross multiplication:

1 mol of K₂CO₃ produce → 1 mol of K₂SO₄, from the stichiometry.

∴ 0.025 mol of K₂CO₃ produce → 0.025 mol of K₂SO₄.

∴ The mass of K₂SO₄ are produced = (no. of moles of K₂SO₄ produced)(molar mass of K₂SO₄) = (0.025 mol)(174.259 g/mol) = 4.35 g.

4. The reaction between zinc and aqueous chromium(III) nitrate can be represented  by the following equation:

3Zn(s) + 2Cr(NO₃)₃ → 3Zn(NO₃)₂ + 2Cr

If 16.25 g of zinc is used to react with chromium(III) nitrate, calculate the mass of  chromium that will be produced.

• Zn reacts with Cr(NO₃)₃ according to the balanced equation:

3Zn(s) + 2Cr(NO₃)₃ → 3Zn(NO₃)₂ + 2Cr ,

It is clear that 3 mole of Zn reacts with 2 mole of Cr(NO₃)₃ to  produce 3 mole of Zn(NO₃)₂  and 2 mole of Cr.

Firstly, we need to calculate the no. of moles of 16.25 g of Zn:

no. of moles of Zn = mass/atomic mass = (16.25 g)/(65.38 g/mol) = 0.25 mol.

using cross multiplication:

3 mol of Zn produce → 2 mol of Cr, from the stichiometry.

∴ 0.25 mol of Zn produce → ??? mol of Cr.

∴ The no. of moles of Cr are produced = (2 mol)(0.25 mol)/(3 mol) = 0.167 mol.

∴ The mass of Cr are produced = (no. of moles of Cr produced)(atomic mass of Cr) = (0.167 mol)(51.9961 g/mol) = 8.67 g.

5. Ethane, C₂H₆, burns in oxygen gas according to the following equation:

2 C₂H₆ + 7 O₂ → 4 CO₂ + 6H₂O,

If 72 dm³ of ethane gas is completely burnt in oxygen, calculate

(a) the volume of carbon dioxide, measured at room temperature and

pressure produced.

Firstly, we can calculate the no. of moles of 72 dm³ ethane at room temperature and pressure using the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.0 atm).

V is the volume of the gas in L (V = 72.0 dm³ = 72.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 298.0 K, room temperature).

∴ n of ethane = PV/RT = (1.0 atm)(72.0 L)/(0.0821 L.atm/mol.K)(298.0 K) = 2.94 mol.

So, we can calculate the no. of moles of CO₂:

using cross multiplication:

2 mol of C₂H₆ produce → 4 mol of CO₂, from the stichiometry.

∴ 2.94 mol of C₂H₆  produce → ??? mol of CO₂.

∴ The no. of moles of CO₂ are produced = (2.94 mol)(4.0 mol)/(2 mol) = 5.88 mol.

∴ The volume of moles of CO₂ are produced = nRT/P = (5.88 mol)(0.0821 L.atm/mol.K)(298.0 K)/(1.0 atm) = 143.86 L.

(b) the volume of oxygen, measured at room temperature and pressure

required​.

using cross multiplication:

2 mol of C₂H₆ require → 7 mol of O₂, from the stichiometry.

∴ 2.94 mol of C₂H₆  require → ??? mol of O₂.

∴ The no. of moles of O₂ are required = (2.94 mol)(7.0 mol)/(2 mol) = 10.29 mol.

∴ The volume of moles of O₂ are produced = nRT/P = (10.29 mol)(0.0821 L.atm/mol.K)(298.0 K)/(1.0 atm) = 251.75 L.