Answer:
SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O
Explanation:
At first calculate the oxidation state of that element which undergoes oxidation as well as reduction.
for SO₄²⁻ the oxidation state of sulphur is +6 and H₂SO₃ the oxidation state of sulphur is +4
So balance equation is
(Reduction) SO₄²⁻ + 4H⁺+ 2e⁻ → H₂SO₃ + H₂O.........................................(1)
(oxidation) Sn²⁺ → Sn⁴⁺ + 2e⁻ .............................................................(2)
Adding equation 1 & 2
we get
SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O
<span>d.2HNO3 (aq) + Sr(OH)2 (aq) → 2H2O (l) + Sr(NO3)2(aq)
4H </span>4H
8O 8O
2N 2N
1Sr 1Sr<span>
</span>
NaHCO3 is the right answer
<span>The calculation of quantities in chemical equations are called Stoichiometry. Stoichiometry is a branch of chemistry which deals with relative quantities of reactants and products in chemical reactions. The correct answer is 'Stoichoimetry'. I hope this helps you. </span>
