Question

2Na+2H2O=2NaOH+H2 In SATP,what is the Volume of H2 if 5g of Na reacts?

Answer 1

We gots `Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr`

Explanation:

And thus HALF an equiv of dihydrogen evolves per equiv of sodium metal...

`"Moles of sodium"=(5.0*g)/(22.99*g*mol^-1)=0.218*mol`

Now at `"SATP"`, `1*mol` of ideal gas occupies `24.8*dm^3`...this specifies (I think) `298*K` and `1*"bar"` (almost an atmosphere)....

And thus ....................

`"volume of dihydrogen"-=24.8*dm^3*mol^-1xx0.218*mol-=5.39*dm^3-=5.39*L`