# Question #71ce2

Nov 9, 2014

The number of moles of Li would be 0.00500 mol and the mass would be 0.0347 g.

There are two reactions taking place.

$2 L i + 2 {H}_{2} O \to 2 L i O H + {H}_{2}$ when the lithium was placed into the water and...

${H}^{+} + O {H}^{-}$$\to {H}_{2} O$ when the acid was added to the resulting solution.

The ${H}^{+}$ and $O {H}^{-}$ react in a 1:1 ratio. This tells us that the number of moles of ${H}^{+}$ used will be equal to the number of $O {H}^{-}$ moles in solution. Likewise, 2 moles of lithium produces 2 moles of $O {H}^{-}$. This is also a 1:1 ratio. As a result, we can say that for every mole of ${H}^{+}$ used from the acid, one mole of lithium must have been added to the water at the start of the reaction.

Now to calculate.

$1 m o l / L \times 0.00500 L = 0.00500 m o l {H}^{+}$
$0.00500 m o l {H}^{+} = 0.00500 m o l L i$

$0.00500 m o l L i \times 6.941 g / m o l = 0.0347 g L i$

Hope this helps.