# (NH4)2Cr2O7 ------> N2 + Cr2O3 + 4H2O if 63 g of ammonium dichromate is heated above 100 degree celcius what will be the loss of mass?

`(NH_4)_2Cr_2O_7 ---------gt N_2 + Cr_2O_3 + 4H_2O`

According to above equation, 1 mol of ammonium dichromate would give  4 mols of `H_2O` . When heated these `H_2O` will be evaporated away.

The molar mass of ammonium dichromate

`= 2 xx(14+4)+ 2xx52 + 7 xx 16`

`= 252 g...

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`(NH_4)_2Cr_2O_7 ---------gt N_2 + Cr_2O_3 + 4H_2O`

According to above equation, 1 mol of ammonium dichromate would give  4 mols of `H_2O` . When heated these `H_2O` will be evaporated away.

The molar mass of ammonium dichromate

`= 2 xx(14+4)+ 2xx52 + 7 xx 16`

`= 252 g per mol`

The number of mols of  ammonium dichromate heated

= 63 g / 252 g per mol

= 0.25 mol.

Therefore number of mols of water evaporated = 4 xx 0.25 mol

= 1 mol.

The mass of water evaporated                  = 1 mol xx 18 g per mol

= 18 g.

Therefore 18 grams of water is evaporated.

Also produced N_2 also are released to atmosphere.

Number of N_2 mols produced                ` = 1 xx 0.25` mol

= 0.25 mol

Mass of N_2 released                              `= (14+14) xx 0.25` g

= 7 g.

Therefore total mass loss           `= 18 + 7`

= 25 g.

Approved by eNotes Editorial Team