Language of Chemistry — Practice Quiz

2Na + 2H₂O → 2NaOH + H₂↑, Na + 2H₂O → NaOH + 2H₂↑

Step 1: Write the skeletal equation: Na + H₂O → NaOH + H₂. Step 2: Count atoms on each side – Reactant: Na=1, H=2, O=1; Product: Na=1, H=3, O=1. Step 3: H is unbalanced. Multiply H₂O and NaOH by 2 on respective sides: 2Na + 2H₂O → 2NaOH + H₂. Step 4: Verify – Na=2, H=4, O=2 on left; Na=2, H=4+2=4? Let's check: 2NaOH gives H=2, H₂ gives H=2, total H=4 ✓, O=2 ✓, Na=2 ✓. Step 5: The upward arrow (↑) shows H₂ is a gas evolved. Option A is the unbalanced skeletal form. Options C and D have incorrect coefficients that violate atom balance.

1, 3, 2, 3

Step 1: Skeletal equation: Fe₂O₃ + CO → Fe + CO₂. Step 2: Count atoms – Reactants: Fe=2, O=3+1=4, C=1; Products: Fe=1, O=2, C=1. Multiple elements are unbalanced. Step 3: Fe₂O₃ has 2 Fe atoms on the left, so we need 2 Fe on the right: Fe₂O₃ + CO → 2Fe + CO₂. Step 4: Now balance oxygen. Left has 3 O from Fe₂O₃ + O from CO; Right has 2 O per CO₂. With 3CO on the left, O from CO = 3, total O = 3+3? No – Fe₂O₃ gives 3 O and 3CO gives 3 O = 6 O total on left... Let us try: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Check: Fe: 2=2 ✓, C: 3=3 ✓, O: 3+3=6 and 3×2=6 ✓. Step 5: Coefficients are 1, 3, 2, 3. Other options

Correct formula is Fe₂(SO₄)₃; student used Fe²⁺ (ferrous) valency instead of Fe³⁺ (ferric)

Step 1: 'Ferric' refers to Iron(III), meaning Fe has valency 3 (Fe³⁺). The student incorrectly used Fe²⁺ (ferrous/Iron II) which gives FeSO₄ (iron(II) sulphate). Step 2: For ferric sulphate: Fe³⁺ and SO₄²⁻. Step 3: Apply criss-cross method – valency of Fe (3) becomes subscript of SO₄, and valency of SO₄ (2) becomes subscript of Fe. Step 4: This gives Fe₂(SO₄)₃. Parentheses are needed around SO₄ since its subscript is 3. Step 5: FeSO₄ is the formula for ferrous sulphate (Iron II sulphate). Confusing ferrous and ferric is a very common mistake in variable valency problems.

The rate at which the reaction proceeds

Step 1: A balanced chemical equation gives qualitative information – names and formulae of substances (reactants and products). Step 2: It gives quantitative information – number of molecules, atoms, and masses of substances involved. Step 3: Molecular masses can be calculated from the formulae using atomic masses, so this information IS available from the equation. Step 4: However, the RATE of reaction (how fast/slow it occurs) is NOT shown in the equation. Rate depends on factors like concentration, temperature, and catalysts. Step 5: This is a key limitation of chemical equations. Other inf