How much dihydrogen would evolve if sodium metal were dropped into water?

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How much dihydrogen would evolve if sodium metal were dropped into water?

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Answer 1 · 2017-07-18T21:08:08

You need to specify a mass of sodium metal.......

$N a (s) + H_{2} O (l) \to N a O H (a q) + \frac{1}{2} H_{2} (g) ↑ ⏐ ⏐ ⏐$ $Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr$

Explanation:

If $1 \cdot g$ $1*g$ of sodium metal were dropped into a beaker of water, along with spitz and sparx we would get......

$\frac{1}{2} \times \frac{1 \cdot g}{22.99 \cdot g \cdot m o l^{- 1}} \times 2.02 \cdot g \cdot m o l^{- 1} = 0.044 \cdot g$ $1/2xx(1*g)/(22.99*g*mol^-1)xx2.02*g*mol^-1=0.044*g$ $dihydrogen gas$ $"dihydrogen gas"$ .

Of course, if we specify the temperature, and assume $1 \cdot a t m$ $1*atm$ pressure, we could calculate the volume of dihydrogen gas produced.

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How much dihydrogen would evolve if sodium metal were dropped into water?

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