Question

In experimental procedure part A (reaction A), what volume, in drops, of 16 M HNO3 is required to react with 0.0115 g of copper (Cu) metal? Assume 20 drops per milliliter. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Answer 1

ANSWER:

Molarity of HNO₃ = 16 M

Mass of copper = 0.0115 g

Atomic mass of copper = 63.5 g/mol

Number of moles of copper = (0.0115 g)(63.5 g/mol)

= 0.00018 mol

20 drops of HNO₃ = 1 mL of HNO₃

Now, from the reaction:

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

we see that,

1 mole of Cu (s) required to react = 4 moles of HNO₃ (aq).

0.00018 moles of Cu (s) required to react = 4 x 0.00018 moles of HNO₃ (aq).

= 0.00072 moles of HNO₃ (aq).

And,

molarity of HNO₃ = (number of moles of HNO₃ x 1000)/(volume of solution in mL)

16 M = (0.00072 mol x 1000)/(volume of solution in mL)

volume of solution = 0.045 mL

Since, 1 mL = 20 drops

0.045 mL = 20 x 0.045 drops = 0.9 drops

Hence, 0.9 drops (approx 1 drop) of 16 M HNO₃ is required to react with 0.0115 g of copper metal.