Verified answer

1 mole of Ba(OH)2 neutralises 2 moles of HNO3

2HNO3(aq) + Ba(OH)2(aq) ----> Ba(NO3)2(aq) + 2H2O(l)

moles HNO3 = molarity x litres

= 5.10x10^-2 M x 0.05105 L

= 0.0026036 mol

1 mole Ba(OH)2 reacts with 2 moles HNO3

Therefore moles Ba(OH)2 = 1/2 x moles HNO3

= 0.0013018 mol

volume (L) = moles / molarity

= 0.0013018 mol / 8.36x10^-2 M

= 0.0156 L

= 15.6 ml (3 sig figs)

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