Dany1111
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how many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia?

Respuesta :

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

sebassandin

Answer:

[tex]m_{(NH_4)_2SO_4}=7928.4g(NH_4)_2SO_4[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2NH_3+H_2SO_4-->(NH_4)_2SO_4[/tex]

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:

[tex]m_{(NH_4)_2SO_4}=60.0molH_2SO_4*\frac{1mol(NH_4)_2SO_4}{1molH_2SO_4} *\frac{132.14g(NH_4)_2SO_4}{1mol(NH_4)_2SO_4} \\m_{(NH_4)_2SO_4}=7928.4g(NH_4)_2SO_4[/tex]

Best regards.

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