Question

How much ca is present in ca(No3)2 that
contains 20g nitrogen?
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Answer 1

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20 g N = 1x 20 /28 = 0.71 mole of Ca(NO3)2.

One molecule of Ca(NO3)2 contains 2 atoms of N.

0.71 mole Ca(NO3)2

= 40 x 0.71

= 28.4 g Ca.

Answer 2

Explanation:

20gn = 1×20/28 =0.71 male of Ca ( No3) 2.

One molecule of Ca (No3) 2 Contains 2 atoms

of N.

0.71 Mole Ca (No3) 2.

= 40 × 0.71

= 28.4g ca