Question

2. From the equation,3Cu + 8HNO3 → 3Cu (NO3)2 + 4H2O + 2NO(At. mass Cu = 64, H = 1, N = 14,0 = 16)calculate :(a) mass of copper needed to react with 63 g of HNO3(b) volume of nitric oxide at STP that can be collected.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

From the equation,3Cu + 8HNO3 → 3Cu (NO3)2 + 4H2O + 2NO(At. mass Cu = 64, H = 1, N = 14,0 = 16). The

(a) mass of copper needed to react with 63 g of HNO3 is 24gm of Cu

(b) volume of nitric oxide at STP that can be collected is ​5.6L

The reaction is as follows:

3Cu + 8HNO3 → 3Cu (NO3)2 + 4H2O + 2NO

Molecular mass of 8 moles of HNO3 is 8(1 + 14 + 3*16)gm = 8(63)gm = 504gm.

Molecular mass of 3 moles of Cu is 3(64) = 192gm.

So, 504gm of HNO3 reacts with 192gm of Cu.

1 gm of HNO3 reacts with (192 ÷ 504) = 0.381gm of Cu.

So, 63 gm of HNO3 reacts with (0.381 × 63) = 24gm of Cu.

(a) Mass of copper needed to react with 63 g of HNO3 is 24gm of Cu.

From 3 moles of Cu (192gm), we get 2 moles of NO gas.

Volume occupied by 2 moles of NO gas = 2(22.4)L = 44.8 L.     [As, volume occupied by 1 mole of any gas at STP is 22.4L].

So, from 192gm of Cu we get 44.8L of NO gas at STP.

And from 24 gm of Cu, we get {[(44.8) ÷ (192)] × 24}L = 5.6L of NO gas.

(b) So, volume of nitric oxide at STP that can be collected is 5.6L