# How much H_2 gas at STP can be produced by the reaction 2 Na(s) + 2 H_2O(l) rightarrow H_2(g) +...

## Question:

How much {eq}H_2 {/eq} gas at {eq}STP {/eq} can be produced by the reaction {eq}2 Na(s) + 2 H_2O(l) \longrightarrow H_2(g) + 2 NaOH (aq) {/eq} of {eq}3.90 \ g {/eq} of {eq}Na {/eq} and excess water ?

## Reaction of Metals with Water and Acids

{eq}{/eq}

1. Reaction of metals with water : Metals react with water to produce metal hydroxides and hydrogen gas. In other words, they reduce the hydrogen in water and bring it to its native state.

$$M + nH_2O \longrightarrow M(OH)_n + \frac{1}{2} H_2 \\$$

Value of n depends on the valency of metal M.

{eq}{/eq}

2. Reaction of metals with acids : Metals react with acids to produce salts and hydrogen gas. In other words, they reduce the hydrogen in the acid and bring it to its native state.

$$M + nHA \longrightarrow MA_n + \frac{1}{2} H_2 \\$$

Note - Metals that are weaker reducing agents than hydrogen ( e.g. Copper, gold, etc) cannot undergo these reactions since they are unable to replace hydrogen from their compounds.

{eq}{/eq}

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{eq}{/eq}

Given chemical equation :

$$\displaystyle 2 Na(s) + 2 H_2O(l) \longrightarrow H_2(g) + 2 NaOH (aq) \\$$

Since water is taken in...