a piece of magnesium ribbon is dissolved in 40ml of N/10HCl completely.the excess of the acid was neutralised by 15ml of N/5 NaOH.the weight of magnesium ribbon is------

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