A piece of Mg is dissolved in 40 ml of N/10 HCL completely. The excess of acid was neutralized by 15 ml of N/5 NaOH. The weight of Mg is(1) 0.24 g(2) 0.024g(3) 0.012g(4) 0.40g

HCl + NaOH→ NaCl + H2O
15mL of N/5 NaOH
N = (n x 1000)/ V
n = 0.003 moles
1 mole of NaOH neutralises 1 mole of HCl
So 0.003 moles of NaOH will neutralise 0.003 moles of HCl

Total number of moles of HCl
40 mL of N/10
n = 0.004 moles

Total number of moles of HCl which reacted with MgCl2= 0.004 - 0.003 = 0.001 moles

Mg + 2 HCl → MgCl2+ H2

2 moles of HCl react with 1 mole of MgCl2

0.001 mole of HCl will react with ( 1/2 ) x 0.001 = 0.0005 moles
Mass of Mg ribbon = 0.0005 x 24 = 0.012 g