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deff fn [24]
3 years ago
6

A 31.3-g sample of ammonium carbonate contains ________ mol of ammonium ions.

Chemistry
2 answers:
Elan Coil [88]3 years ago
8 0

Answer:

n_{NH_4}=0.652molNH_4

Explanation:

Hello,

Ammonium carbonate has the following formula:

(NH_4)_2CO_3

In this manner, the moles are computed by knowing its molecular mass:

M_{(NH_4)_2CO_3}=14*2+1*8+12*1+16*3=96g/mol

Thus, by applying a mass-mole-ions relationship, one obtains:

ions_{(NH_4)_2CO_3}=31.3g(NH_4)_2CO_3*\frac{1mol(NH_4)_2CO_3}{96g(NH_4)_2CO_3}*\frac{2molNH_4}{1mol(NH_4)_2CO_3}\\n_{NH_4}=0.652molNH_4

Best regards.

Leona [35]3 years ago
7 0
<span>We need to calculate the equivalent amount in units of moles of ammonium ions from the mass units. For this we need the molar mass of the substances involved. We calculate as follows: 

31.3 g </span>(NH4)2CO3 ( 1 mol (NH4)2CO3 / 96.09 g (NH4)2CO3) ( 2 mol NH4 / 1 mol (NH4)2CO3 ) = 0.65 mol <span>ammonium ions</span>
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