how many mL of 3.0 M NaOH are required to react with 4.0 mL of 16M HNO3?

Question

this has to do with the cycle of copper reactions experiment

Answer 1

In a neutralization reaction, the moles of acid must equal the moles of base.

This is easy found by M(acid) x V(acid) = M(base) x V(base)

So, 4.0 x 16 = 3.0 x V(base)

64 = 3.0 x V(base)

V(base) = 21 mL