Question

Sr(OH)2 (aq) + 2 HF (aq) --> 2 H2O (l) + SrF2 (l)





If 345 mL of 5.25 M Sr(OH)2 is used, how many mL of 7.34 M HF can be neutralized with it?

Answer 1

$n=C*V$

$n(Sr(OH)_2)=5.25M*0.345L=1.70625 mol$

According to the chemical reaction equation

$n(HF)=1.70625mol*2=3.4125 mol$

$V(HF)=3.4125mol/7.34M=0.465L$

The answer is 465 ml