Question

What mass of dihydrogen results from the reaction of a `0.4*mol` quantity of sodium metal with water?

What mass of dihydrogen results from the reaction of a `0.4*mol` quantity of sodium metal with water?

Answer 1

How many eggs do you use when you make scrambled eggs from a dozen eggs?

Explanation:

And this is precisely the same sort of question.

One mole specifies `"Avogadro's number"` of particles, i.e. `6.022xx10^23` particles. Why should we use such an absurdly large number? Well `"Avogadro's number"` of `""^1H` atoms has a mass of `1.00*g` precisely; `"Avogadro's number"`, `"one mole"` of `Na` atoms has a mass of `22.99*g` precisely.

The molar mass of the all the other elements are listed on the Periodic Table. And `"Avogadro's number"` is thus the link between grams, and kilograms, and litres, that which we measure on a bench, with the sub-micro world of atoms, and molecules, of which we CONCEIVE.

Now from the table, `0.4*mol*Na-=0.4*molxx22.99*g*mol^-1=9.20*g` with respect to sodium .... it reacts according to the equation.....

`Na(s)+H_2O(l)rarrNaOH(aq) + 1/2H_2(g)uarr`

We get `0.2*mol` dihydrogen gas, i.e. `0.40*g`, and `0.4*mol` `NaOH(aq)`. What is the mass of water that reacts?

And please note the dimensional consistency of the answer. We want a mass, an answer in grams..

`0.4*mol*Na-=0.4*cancel(mol)xx22.99*g*cancel(mol^-1)=9.20*g`

The product gives us an answer with units of grams, which persuades us that the we got the order of operations right. It is all too easy to divide instead of multiply or vice versa.....this use of units is an example of dimensional analysis.