Question #93f99

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Answer 1 · 2014-11-30T23:17:55

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 $HNO_3= 0.01434xx2= 0.02868$

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