Question #cb30d

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Question #cb30d

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Answer 1 · 2017-06-02T16:50:55

165.6g

Explanation:

Number of moles of $N a N O_{3}$ $NaNO_3$ in 85g = 85/85 = 1 mole

$P b O + 2 N a N O_{3} \to P b {(N O_{3})}_{2} + N a_{2} O$ $PbO + 2NaNO_3 rarr Pb(NO_3)_2 + Na_2O$

2 moles of $N a N O_{3}$ $NaNO_3$ required to react with $P b O$ $PbO$ to form $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ . Therefore, 0.5 moles of $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ are produced.
Molar mass of $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ = 331.2g
Therefore, mass of $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ produced = 331.2/2 = 165.6g

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