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Double Displacement Reaction

PbNO32+Na2SO4...

Question

$P b (N O_{3})_{2} + N a_{2} S O_{4} ⟶$

2 N a N O_{3} + P b S O_{3}

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N a N O_{3} + P b S O_{4}

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2 N a N O_{3} + P b S O_{4}

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2 N a N O_{2} + P b S O_{4}

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Solution

The correct option is D $2 N a N O_{3} + P b S O_{4}$
$P b (N O_{3})_{2} + N a_{2} S O_{4} ⟶ 2 N a N O_{3} + P b S O_{4}$
It is a kind of double displacement reaction.

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Similar questions

When excess lead nitrate solution was added to a solution of sodium sulphate, 15.15 g of lead sulphate were precipitated.

What mass of sodium sulphate was present in the original solution?

Na₂SO₄ + Pb(NO₃)₂ → PbSO₄ + 2NaNO₃

[Na = 23, S = 32, O = 16, Pb = 207, N = 14]

P b S O_{4}

N a_{2} S O_{4}

K_{s p}

P b S O_{4} = 1.25 \times 10^{- 9}

N a C l, N a {N O}_{3}, {({N H}_{4})}_{2} {S O}_{4}, P b {S O}_{4}

O_{2}

N a N O_{3}

2 N a N O_{3} \to 2 N a N O_{2} + O_{2}

P b S O_{4}

20.0

0.1

P b (N O_{3})_{2}

30.0

0.1

N a_{2} S O_{4}

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