Question

Question #79474

Answer 1

A bit under `5*g` of ammonia will be produced.

Explanation:

`(NH_4)_2SO_4(aq) + Ca(OH)_2(aq) rarr 2NH_3(aq) + CaSO_4(s)darr + 2H_2O`

`"Moles of "Ca(OH)_2` `=` `(10.0*g)/(74.09*g*mol^-1)=0.135*mol`

And thus, given the stoichiometry, `0.270*mol` of ammonia will be generated.

This has a mass of `0.270*molxx17.03*g*mol^-1=??g`