Question

Q6.
Copper metal reacts with nitric acid according to the reaction (Cu = 63.5 u)
3Cu +8HNO3 → 3Cu(NO3)2 + 2NO + 4H20
If 9.375 g copper nitrate is eventually formed, how many g of NO are formed?​

Answer 1

Explanation:

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Answer 2

Given:

reaction involved = 3Cu +8HNO3 → 3Cu(NO3)2 + 2NO + 4H20

Cu(NO3)2 formed = 9.375 g

To Find:

how many g of NO are formed =?

Explanation:

Now, in order to find limiting reagent,

from reaction,  3Cu +8HNO3 → 3Cu(NO3)2 + 2NO + 4H20

we infer that,

3 mol Cu → 3 mol Cu(NO3)2

then 1 mol Cu → 1 mol Cu(NO3)2

Also, 8 mol HNO3 → 3mol Cu(NO3)2

then 1 mol HNO3 → 3/8 mol Cu(NO3)2

therefore HNO3 will act as a limiting reagent.

Now, molar mass of HNO3 = 63 g/mol

         molar mass of Cu(NO3)2 = 187.5 g/mol

         molar mass of NO =  30 g/mol

1 mol HNO3 → 3/8 mol Cu(NO3)2

63g HNO3 → 3/8 × 187.5 g Cu(NO3)2

9.375g Cu(NO3)2  → $\dfrac{63\times8\times 9.375}{3\times 187.5}$ g HNO3

Now, from the reaction, we also know that

1 mol HNO3 → 2 mol NO

$\dfrac{63\times8\times 9.375}{3\times 187.5}$ g HNO3 →  $\dfrac{2\times30\times 9.375\times63\times8}{63\times 187.5\times3}$ g NO

$\dfrac{2\times30\times 9.375\times63\times8}{63\times 187.5\times3}$   = 7.99g NO

Thus approximately 8 g NO is formed