Question

Question #93f99

Answer 1

The molarity of the acid is 1.434.

Start with the equation:

`2HNO_(3(aq))+Ca(OH)_(2(aq))rarrCa(NO_3)_(2(aq))+2H_2O_((l))`

This tells us that 2 moles of `HNO_3` reacts with 1 mole of `Ca(OH)_2`

`c=(n)/(v)`
So `n=cv`

So no. moles `Ca(OH)_2=(17.42xx0.823)/(1000)=0.01434`

So no. moles `HNO_3= 0.01434xx2= 0.02868`

`c=(n)/(v)=(0.02868)/(0.02)=1.434mol.dm^(-3)`