What is the resulting volume given that ten cm3 of a two-molar solution of nitric acid is added to ten cm3 of a one-molar solution of calcium hydroxide?

Answer 1

If you are being literal, then the volume might actually increase slightly because of the formation of H₂O during the neutralization reaction: 2HNO₃ + Ca(OH)₂ ==> CaCl₂ + 2H₂O

Looking at the values:

moles HNO₃ = 0.01 L x 2 mol/L = 0.02 moles HNO₃

moles Ca(OH)₂ = 0.01 L x 1 mol/L = 0.01 moles Ca(OH)₂

Neither is limiting so 0.02 moles H₂O should be formed

0.02 moles H₂O x 18 g/mole = 0.36 g H₂O

Assuming a density of 1g/ml you'd have an additional 0.36 mls of volume

So, "theoretically", the final volume would be 10 ml + 10 ml + 0.36 ml = 20.36 mls.

HOWEVER, a large amount of heat is also generated in this reaction, and the density of H₂O cannot be taken as 1g/ml, so that's another compounding factor.

I'd say that this question can't really be answered completely without significantly more information, but if it was meant to be a exercise in balancing the equation and calculating the water formed, then the above explanation might be suitable.

Answer 2

Given the conversion factor, 1 cm³ is equal to 0.001L or 1 mL, to find the total volume you will add both volumes together.

10 cm³ + 10 c m³ = 20 cm^3.

^{Thus the volume can be 20 mL or 0.020 L.}