What volume of 0.436 M barium nitrate solution is needed to prepare 240.0 mL of a barium nitrate solution that is 0.300 M in nit
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Answer:
Explanation:
NO₃⁻ + Sn²⁺ + __ → NO₂ + Sn⁴⁺ + __
Step 1: Separate into two half-reactions.
NO₃⁻ ⟶ NO₂
Sn²⁺ ⟶ Sn⁴⁺
Step 2: Balance all atoms other than H and O.
Done
Step 3: Balance O.
NO₃⁻ ⟶ NO₂ + H₂O
Sn²⁺ ⟶ Sn⁴⁺
Step 4: Balance H
NO₃⁻ + 2H⁺ ⟶ NO₂ + H₂O
Sn²⁺ ⟶ Sn⁴⁺
Step 5: Balance charge.
NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O
Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻
Step 6: Equalize electrons transferred.
2 × [NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O]
1 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]
Step 7: Add the two half-reactions.
2 × [NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O]
<u>1 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻] </u>
2NO₃⁻ + Sn²⁺ + 4H⁺ ⟶ 2NO₂ + Sn⁴⁺ + 2H₂O
Step 8: Check mass balance.
On the left: 2 N, 6 O, 1 Sn, 4H
On the right: 2 N, 6 O, 1 Sn, 4H
Step 9: Check charge balance.
On the left: -2 + 6 = +4
On the right: +4
The equation is balanced.