Question #198a3

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Question #198a3

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Answer 1 · 2017-05-11T21:14:16

We exploit the stoichiometric reaction............

Explanation:

$B a {(N O_{3})}_{2} (a q) + N a_{2} C O_{3} (a q) \to B a C O_{3} (s) ⏐ ↓ + 2 N a N O_{3} (a q)$ $Ba(NO_3)_2(aq) + Na_2CO_3(aq) rarr BaCO_3(s)darr+2NaNO_3(aq)$

Most carbonates are insoluble; the alkaline earth carbonates are especially insoluble. $B a C O_{3}$ $BaCO_3$ has $K_{s p} = 2.58 \times 10^{- 9}$ $K_(sp)=2.58xx10^-9$ (these data should have been supplied with the question).

Addition of stoichiometric sodium carbonate should suffice.

$Moles of barium nitrate = 120 \times 10^{- 3} \cdot L \times 0.0500 \cdot m o l \cdot L^{- 1} = 6 \times 10^{- 3} \cdot m o l .$ $"Moles of barium nitrate"=120xx10^-3*Lxx0.0500*mol*L^-1=6xx10^-3*mol.$

And thus we need.....................................................................

$6 \times 10^{- 3} \cdot m o l \times 105.99 \cdot g \cdot m o l^{- 1} = 635 \cdot m g$ $6xx10^-3*molxx105.99*g*mol^-1=635*mg$ with respect to $sodium carbonate$ $"sodium carbonate"$ .

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Question #198a3

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