Calculate the volume (in mL) of 0.416 M HNO3 needed to react completely with 8.18 g of ZnCO3 in a gas-forming reaction?

Question

I don't understand how to solve this. Please help me with this problem.

Answer 1

The first thing you must do is write the correctly balanced equation for this reaction:

2HNO₃(aq) + ZnCO₃(s) ==> H₂CO₃(aq) + Zn(NO₃)₂(aq) ==> CO₂(g) + H₂O(l) + Zn(NO₃)₂(aq)

Next, we find the MOLES of ZnCO₃ represented by 8.18 g (MW ZnCO3 = 125 g/mol):

8.18 g ZnCO₃ x 1 mole/125 g = 0.06544 moles

Using the stoichiometry of the balanced equation (2 HNO₃ : 1 ZnCO₃) we find MOLES of HNO₃ present:

0.06544 moles ZnCO₃ x 2 moles HNO₃/1 mole ZnCO₃ = 0.13088 moles

Finally, we calculate the volume (in ml) of 0.416 M (0.416 mol/L) of HNO₃ needed to provide 0.13088 moles:

(x L)(0.416 moles/L)(1000 ml/L) = 0.13088 moles and x = 315 mls of HNO₃ needed (to 3 significant figures)