Question

Cu + 4 HNO3 —-> Cu(NO3)2 + 2 NO2 + 2 H2O

Given the mass of .5g of Copper measured; calculate the minimum mL of 6M HNO3! Required to fully react?

Answer 1

1 mole copper reacts with 4 mole nitric acid

So

63 gm of copper will reacts with $4\times 63=252 \ gm$ nitric acid

So 5 gm copper will required 20 gm Nitric acid

Now calculating the amount of nitric acid required:

$$Molarity = \frac{moles }{volume }= \frac{20\times1000}{63\times vol(ml)}=6 \\vol =\frac{20000}{378}=52.91ml$$