1. A. 9 eggs
2. B. 5:3
3. C. Exactly 44 g
4. C. Mass and atoms only
5. D. 6.81 g of PH₃
6. C. 88.4%
7. B. O₂
8. D. 30.8 g CO₂
9. B.
10. B.
11. C. 99 g
12. D. mole ratio
13. C. Theoretical
14. A. 6.0 mol H₂O
15. D. 39.7 g CH₃OH
16. A. 650 g HgO
17. D. 8.8 mol H₂
18. B. 82.6%
Explanation:
1. The number of eggs it takes to make 1 cake = 3 eggs
The number of eggs it takes to make 3 × 1 = 3 cake = 3 × 3 = 9 eggs
Therefore, the correct option is;
A. 9 eggs
2. The given reaction is presented as follows;
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
In the above reaction, we have;
Moles of , O₂, reacted = 5 moles
Moles of , CO₂, produced = 3 moles
The ratio of the of O₂ reacted to moles of CO₂ produced = 5 moles:3 moles
∴ The ratio of the of O₂ reacted to moles of CO₂ produced = 5:3
The correct option is;
B. 5:3
3. The reaction is presented as follows;
O₂ (g) + C(s) → CO₂ (g)
From the reaction, 1 mole (12 g) of carbon produces 1 mole of CO₂
The molar mass of CO₂ = The mass of 1 mole of CO₂ = 44.01 g/mol
Given that the reaction is completed, the mass of CO₂ produced = The mass of 1 mole of CO₂ ≈ 44 g
The correct option is;
C. Exactly 44 g
4. The given reaction is presented as follows;
N₂ + 3 F₂ → 2NF₃
The initial number of atom = 2 + 6 = 8
The final number of atom = 2 × 4 = 8
∴ The initial number of atom = The final number of atom
Therefore, the number of atoms is conserved;
The mass of the reactants ≈ 28 g/mol + 3 × 37.996806 g/mol ≈ 141.993612 g/mol ≈ 142 g/mol
The mass of the product ≈ 2 × 71 g.mol = 142 g/mol
∴ The mass is conserved
Moles of reactants = 1 + 3 = 4
Moles of products = 2
∴ The number of moles is not conserved
The correct option is
C. Mass and atoms only
5. The molar mass of P₄ = 123.895048 g/mol
One mole of P₄ (123.895048 g) produces four moles (4 × 34.00) of PH₃
6.20 g of P₄. will produce (4 × 34.00)/(123.895048) × 6.20 g ≈ 6.80576 g ≈ 6.81 g
The correct option is D. 6.81 g of PH₃
6. The percentage yield = ((The actual yield)/(The ideal yield)) × 100
The actual yield of silver = 38.1 g
The ideal yield of silver = 43.1 g
∴ The percentage yield = ((38.1 g)/(43.1 g)) × 100 = 88.3990719258% ≈ 88.4%
The percentage yield = 88.4%
The correct option is C. 88.4%
7. The given chemical equation is presented as follows;
CS₂ (g) + 3 O₂ (g) → CO₂ (g) + 2 SO₂ (g)
The number of moles in 192 g of O₂ = 192 g/(32 g/mol) = 6 moles
Given that 3 moles of O₂ reacts with 1 mole of CS₂ to produce 1 mole of CO₂ and 2 moles SO₂, therefore 2 × 3 = 6 moles of O₂ will reacts with 2 × 1 = 2 moles of CS₂ to produce 2 moles of CO₂ and 4 moles SO₂
∴ The limiting reactant is;
B. O₂
8. The given chemical equation is presented as follows;
2 C₈H₁₈ (g) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O (l)
The number of moles in 10 g of C₈H₁₈, n₁ = (10 g)/(114.26 g/mol)
The number of moles of CO₂ produced, n₂ = (10 g)/(114.26 g/mol) × 16/2 ≈ 0.7 moles
The mass of CO₂ produced, m ≈ 44.01 × n₂ ≈ 44.01 g/mol × 0.7 moles ≈ 30.807 grams ≈ 30.8 grams
The theoretical yield of CO₂ from completely burning 10.0 g of C₈H₁₈ ≈ 30.8 grams of CO₂
The correct option is D. 30.8 g CO₂
9. The correct option is B. The limiting reactants determine the maximum amount of product that can be formed
10. Option B, which has 3 atoms of each element combining to form a product with 1 atom of one element and 2 atoms of the other element
The correct option is B.
11. By the conservation of mass, we have;
The mass of the reactants = The mass of the products
Let 'x' represent the mass of zinc in the products of the reaction
Therefore, we have;
61 g of calcium + 207 g of zinc chloride = 169 g of calcium chloride + x g of Zinc
∴ x g = 61 g + 207 g - 169 g = 99 g
The mass of zinc in the products of the reaction, x g = 99 g
The correct option is;
C. 99 g
12. The quantity that must be used to convert from one chemical substance to another is the mole ratio
The correct option is D. mole ratio
13. The maximum mass of the product that could form in a reaction is called the theoretical yield, which is option C.
The correct option is C. Theoretical
14. 1 mole of O₂ produces 2 moles of water (H₂O), therefore;
3 × 1 = 3.0 moles of O₂ will produce 3 × 2 = 6 moles of H₂O
The correct option is
A. 6.0 mol H₂O
15. 2 mole × 2.02 g/mol = 4.04 g of H₂ (g) produces 32.05 g CH₃OH (l)
∴ 5 g of H₂ (g) will produce 32.05 g × 5/4.04 ≈ 39.6658416 g ≈ 39.7 g of CH₃OH
The correct option is;
D. 39.7 g CH₃OH
16. 2 (2 × 216.59 g = 433.18 g) moles of HgO produces 1 mole of O₂
1.5 mole of O₂ will be produced by 1.5 × 2 = 3 moles (3 × 216.59 g = 649.77 g ≈ 650 g) of HgO
The correct option is A 650 g HgO
17. 3 moles of H₂ produces 2 moles of NH₃
The number of moles of NH₃ in 100 g of NH₃, n = 100 g/(17.04 g/mol) = 5.868544 moles
The number of moles of H₂ that will produce 5.868544 moles of NH₃ = 3/2 × 5.868544 moles = 8.802816 moles ≈ 8.8 moles
Therefore, the correct option is;
D. 8.8 mol H₂
18. The theoretical yield of PbO = (223.2/331.2) × 9.90 g = 6.67173913 g
The percentage yield = (5.51 g)/(6.67173913 g) × 100 ≈ 82.6%
The correct option is option B 82.6%.
