Question

how many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia?

Answer 1

2NH3+H2SO4=>(NH4)2SO4
60(1/1)=60mol (NH4)2SO4
1mol=132.14grams
=7928.4grams

Answer 2

`m_((NH_4)_2SO_4)=7928.4g(NH_4)_2SO_4`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`2NH_3+H_2SO_4-->(NH_4)_2SO_4`

Now, by taking into account that 60.0 mol of sulfuric acid are used, the produced grams of ammonium sulfate, by applying the stoichiometric factors, turn out as shown below:

`m_((NH_4)_2SO_4)=60.0molH_2SO_4*(1mol(NH_4)_2SO_4)/(1molH_2SO_4) *(132.14g(NH_4)_2SO_4)/(1mol(NH_4)_2SO_4) \\m_((NH_4)_2SO_4)=7928.4g(NH_4)_2SO_4`

Best regards.