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------

Asked by ashwinivg | 4th Jun, 2010, 12:00: AM

Expert Answer:

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

Answered by  | 22nd Jul, 2010, 11:52: AM

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