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3 years ago

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

Chemistry

2 answers:

kkurt [141]3 years ago

7 0

Answer:

$m_{(NH_4)_2SO_4}=7928.4g(NH_4)_2SO_4$

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:

$m_{(NH_4)_2SO_4}=60.0molH_2SO_4*\frac{1mol(NH_4)_2SO_4}{1molH_2SO_4} *\frac{132.14g(NH_4)_2SO_4}{1mol(NH_4)_2SO_4} \\m_{(NH_4)_2SO_4}=7928.4g(NH_4)_2SO_4$

Best regards.

lianna [129]3 years ago

5 0

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

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Carbon: [He] $2s^{2} 2p^{2}$

Fluorine: [He] $2s^{2} 2p^{5}$

Hydrogen: $1s^{1}$

Nitrogen: [He] $2s^{2} 2p^{3}$

Oxygen: [He] $2s^{2} 2p^{4}$

Sodium: [Ne] $3s^{1}$

The period of a chemical element is given by the last energy level of electron configuration.

The group of a chemical element is given by the amount of electrons in the last energy level of electron configuration.

Therefore,

Carbon: Period 2 Group 4

Fluorine: Period 2 Group 7

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