# For the reaction 2Na + 2H_2O -> NaOH + H_2, how many grams of hydrogen are produced from 130 g of sodium?

May 25, 2017

Well how many moles of sodium metal do you have......?

#### Explanation:

$\text{Moles of sodium} = \frac{130 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 5.65 \cdot m o l .$

Now you have the stoichiometric equation:

$2 N a \left(s\right) + 2 {H}_{2} O \left(l\right) \rightarrow 2 N a O H \left(a q\right) + {H}_{2} \left(g\right) \uparrow$

If there are $5.65 \cdot m o l$ of metal, clearly $\frac{5.65 \cdot m o l}{2}$ dihydrogen gas are evolved......, i.e. $2.83 \cdot m o l$ ${H}_{2}$. Typically, in a question like this, you would also be asked to calculate the volume the gas occupies under standard conditions, and the $p H$ of the resultant solution (would it be high, low, neutral?).

Just as an aside, I would not want to be the mug who placed such a quantity of sodium metal in water. The reaction would be violent and explosive.