How many milliliters of 0.200M $N a O H$ $NaOH$ solution are needed to react with 22.0 mL of a 0.490 M $N i C l_{2}$ $NiCl_2$ solution?

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How many milliliters of 0.200M $N a O H$ $NaOH$ solution are needed to react with 22.0 mL of a 0.490 M $N i C l_{2}$ $NiCl_2$ solution?

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Answer 1 · 2015-12-08T18:02:32

There are $0.0108$ $0.0108$ $m o l$ $mol$ $N i C l_{2}$ $NiCl_2$ . $0.0216$ $0.0216$ $m o l$ $mol$ $N a O H$ $NaOH$ are required, given $0.200$ $0.200$ $m o l \cdot L^{- 1}$ $mol*L^-1$ $N a O H (a q)$ $NaOH(aq)$ .

Explanation:

$N i C l_{2} (a q) + 2 N a O H (a q) \to N i {(O H)}_{2} (s) ⏐ ↓ + 2 N a C l (a q)$ $NiCl_2(aq) + 2NaOH(aq) rarr Ni(OH)_2(s)darr + 2NaCl(aq)$

There are $0.0108$ $0.0108$ $m o l$ $mol$ $N i C l_{2}$ $NiCl_2$ ( $22.0xx10^(-3)cancelLxx0.490*mol*cancel(L^-1)$ ). So, by stoichiometry $0.0216$ $mol$ $NaOH$ are required.

There is $0.200$ $mol*L^-1$ $NaOH$ solution available, so...

$(0.0216*cancel(mol))/(0.200*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)$ $=$ $?? mL$

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How many milliliters of 0.200M $N a O H$ $NaOH$ solution are needed to react with 22.0 mL of a 0.490 M $N i C l_{2}$ $NiCl_2$ solution?

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How many milliliters of 0.200M NaOHNaOHNaOH solution are needed to react with 22.0 mL of a 0.490 M NiCl_2NiCl2NiCl_2 solution?

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How many milliliters of 0.200M $N a O H$ $NaOH$ solution are needed to react with 22.0 mL of a 0.490 M $N i C l_{2}$ $NiCl_2$ solution?