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:
<em>2. Iron reacts with oxygen gas according to the following equation:
</em>
<em>4Fe + 3O₂ → 2Fe₂O₃
, If 2 moles of oxygen gas is used in the reaction,
</em>
- Fe reacts with O₂ according to the balanced equation:
<em>4Fe + 3O₂ → 2Fe₂O₃,</em>
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?
<u><em>using cross multiplication:
</em></u>
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?
<u><em>using cross multiplication:
</em></u>
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.
<em>3. Potassium sulfate can be prepared by the reaction between dilute sulfuric acid and potassium carbonate.
</em>
<em>H₂SO₄ + K₂CO₃ → K₂SO₄ + CO₂ + H₂O,
</em>
<em>Calculate the mass of potassium sulfate that can be prepared from 3.45 g of potassium carbonate.</em>
- H₂SO₄ reacts with K₂CO₃ according to the balanced equation:
H₂SO₄ + K₂CO₃ → K₂SO₄ + CO₂ + H₂O<em>,</em>
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.
<u><em>using cross multiplication:
</em></u>
1 mol of K₂CO₃ produce → 1 mol of K₂SO₄, from the stichiometry.
∴ 0.025 mol of K₂CO₃ produce → <em>0.025 mol of K₂SO₄.</em>
∴ 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
<em>,</em>
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.
<u><em>using cross multiplication:
</em></u>
3 mol of Zn produce → 2 mol of Cr, from the stichiometry.
∴ 0.25 mol of Zn produce → <em>??? mol of Cr.</em>
∴ 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.
<em>5. Ethane, C₂H₆, burns in oxygen gas according to the following equation:
</em>
<em>2 C₂H₆ + 7 O₂ → 4 CO₂ + 6H₂O,
</em>
<em>If 72 dm³ of ethane gas is completely burnt in oxygen, calculate
</em>
(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:
<u><em>using cross multiplication:
</em></u>
2 mol of C₂H₆ produce → 4 mol of CO₂, from the stichiometry.
∴ 2.94 mol of C₂H₆ produce → <em>??? mol of </em>CO₂<em>.</em>
∴ 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.
<u><em>using cross multiplication:
</em></u>
2 mol of C₂H₆ require → 7 mol of O₂, from the stichiometry.
∴ 2.94 mol of C₂H₆ require → <em>??? mol of </em>O₂<em>.</em>
∴ 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.
