Home
Class 12
CHEMISTRY
A sample of magnesium was burnt in air t...

A sample of magnesium was burnt in air to give a mixture of MgO and `Mg_3N_2`. The ash was dissolved in 60 m" Eq of "HCl and the resulting solution was back titrated with NaOH. 12 m" Eq of "NaOH was required to reach the end point. An excess of NaOH was then added and the solution distilled. The ammonia released was then trapped in 10 m" Eq of "a second acid solution. Back titration to this acid solution required 6 m" Eq of "the base. Calculate the percentage of magnesium burnt to the nitride.

Text Solution

Verified by Experts

Total meq. of HCl = 60
Unreacted Meq. of HCl = meq of NaOH
Unreacted meq. of HCl = 12
Used meq. of HCl = 60 -12 = 48
Used meq. of HCl = No. of meq. of MgO and `Mg_(3) N_(2)`
`2Mg+ O_(2) rarr 2MgO`
( from air )
`MgO+ 2HCl rarr MgCl_(2) + H_(2)O`
But `Mg_(3) N_(2) + 8 HCl rarr 3 MgCl_(2) + 2NH_(4) Cl`
`NH_(4)Cl + NaOH rarr NaCl + H_(2) O + NH_(3)`
NO. of meq. of `NH_(4) Cl -= ` No. of Meq. of `NH_(3)`
`= ( 10 - 6) = 4 `
No. of meq. of `Mg_(3) N_(2) = ( 1)/( 2)` [ No. of millimole of `NH_(4) Cl] = ( 4)/( 2) = 2 `
No. of millimole ( or meq. ) of HCl used by `Mg_(3) N_(2) = 2 xx 8 = 16`
Thus, no. of meq. of HCl used by `MgO = 48 - 16 = 32 `
No. of millimoles of `MgO = ( 32)/( 2) = 16`
No. of meq. of MgO `= 16 xx 2 = 32 `
No. of meq. of Mg burtn to MgO = 32
`:.` Weight of Mg burnt to MgO `= ( 32 xx 12 )/(1000) = 0.389g `
From equation, `3Mg + N_(2) hArr Mg_(3) N_(2)`
2 millimoles of `Mg_(3) N_(2) = 6` millimoles of Mg
Weight of Mg burnt to `Mg_(3) N_(2) = ( 6 xx 24)/ ( 1000) = 0.144 g `
Total weight of Mg `= 0.384 + 0.144 = 0.528g `
% of Mg burtn of `Mg_(3) N_(2) = ( 0.144 xx 100) /( 0.528 ) = 27.27 %`
Promotional Banner

Similar Questions

Explore conceptually related problems

A sample of Mg was burnt in air to give a mixure of MgO and Mg_(3)N_(2) . The ash was dissolved in 60 Meq . of HCl and the resulting solution was back titrated with NaOH . 12 Meq . Of NaOH was then added and the solution distrilled. The ammonia released was then trapped in 10 Meq . of second acid solution. Back titration of this solution required 6 Meq . of the base Calculate the percentage of Mg burnt to the nitride.

20 gm of sample Ba(OH)_(2) is dissolved in 10 ml of 0.5 N HCl solution, The excess of HCl was titrated with 0.2N NaOH. The volume of NaOH used was 10 ml. Calculate the % of Ba(OH)_(2) in the sample.

When 100 ml of M/10 NaOH solution and 50 ml of M/5 HCI solution are mixed, the pH of resulting solution would be

100ml of 0.2 M H_(2)SO_(4) is added to 100 ml of 0.2 M NaOH . The resulting solution will be

200 " mL of " a solution of a mixture of NaOH and Na_2CO_3 was first titrated with 0.1 M HCl using phenolphthalein indicator. 17.5 " mL of " HCl was required for the same HCl was again required for next end point. Find the amount of NaOH and Na_2CO_3 in the mixture.

2.5 litre of 1 M NaOH solution are mixed with another 3 litre of 0.5 M NaOH solution Then the molarity of the resultant solution is

When 20 mL of M//20NaOH is added to 10mL of M//10 HCI , the resulting solution will

What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?

40 mg of pure NaOH is dissolved in 10 litres of distilled water. The pH of the solution is

A sample of pure sodium carbonate 0.318g is dissolved in water and titrated with HCl solution. A volume of 60mL is required to reach the methyl orange end point. Calculate the molarity of the acid.