What mass of $M {g (O H)}_{2}$ $Mg(OH)_2$ is produced when 100 mL of 0.42 M $M {g (N O_{3})}_{2}$ $Mg(NO_3)_2$ is added to excess $N a O H$ $NaOH$ solution?

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What mass of $M {g (O H)}_{2}$ $Mg(OH)_2$ is produced when 100 mL of 0.42 M $M {g (N O_{3})}_{2}$ $Mg(NO_3)_2$ is added to excess $N a O H$ $NaOH$ solution?

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Answer 1 · 2017-02-19T16:30:21

Mole ratio for the reaction

$M {g (N O 3)}_{2} + 2 N a O H = 2 N a N O_{3} + M {g (O H)}_{2}$ $Mg(NO3)_2 + 2NaOH = 2NaNO_3 + Mg(OH)_2$

1 : 2 :: 2 : 1

For this we need to calculate number of moles in 100ml
of 0.42 M

$\frac{100}{1000} \cdot 0.42 = 0.042$ $100/1000 * 0.42 = 0.042$

$0.042 : 0.042 \cdot 2 : : 0.042 \cdot 2 : 0.042$ $0.042 : 0.042*2 :: 0.042 *2 : 0.042$

$0.042 m o l M {g (N O 3)}_{2} : 0.084molNaOH : : 0.084mol N a N O_{3} : 0.042 m o l M {g (O H)}_{2}$ $0.042mol Mg(NO3)_2 : "0.084molNaOH" :: "0.084mol "NaNO_3 : 0.042mol Mg(OH)_2"$

0.042 moles of $M {g (O H)}_{2}$ $Mg(OH)_2$ are formed

$Mass = molar mass \times moles$ $"Mass" = "molar mass" xx "moles"$

Molar mass of $M {g (O H)}_{2}$ $Mg(OH)_2$ = 58.3197

Mass of $M {g (O H)}_{2}$ $Mg(OH)_2$ formed = $58.3197 \times 0.042$ $58.3197 xx 0.042$

2.4494274grams of $M {g (O H)}_{2}$ $Mg(OH)_2$

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What mass of $M {g (O H)}_{2}$ $Mg(OH)_2$ is produced when 100 mL of 0.42 M $M {g (N O_{3})}_{2}$ $Mg(NO_3)_2$ is added to excess $N a O H$ $NaOH$ solution?

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What mass of $M {g (O H)}_{2}$ $Mg(OH)_2$ is produced when 100 mL of 0.42 M $M {g (N O_{3})}_{2}$ $Mg(NO_3)_2$ is added to excess $N a O H$ $NaOH$ solution?