$(N H_{4})_{2} C r_{2} O_{7} \to C r_{2} O_{3} + 4 H_{2} O + N_{2}$
Calculate the volume of nitrogen at STP, evolved when 63 g of ammonium dichromate is heated.

5.6 litre

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11.2litre

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22.4 litre

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18.4 litre

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Solution

The correct option is A 5.6 litre
$(N H_{4})_{2} C r_{2} O_{7} \to C r_{2} O_{3} + 4 H_{2} O + N_{2}$
One mole of ammonium dichromate will release one mole of nitrogen gas.
252 gm of ammonium dichromate will release 22.4 litre nitrogen gas.
So, 63 gm will give $22.4 \times \frac{63}{252} = \frac{22.4}{4} = 5.6 L N_{2}$ .

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Similar questions

(NH₄)₂Cr₂O₇ $\overset{}{\to}$ N₂ + Cr₂O₃ + 4H₂O. What volume of nitrogen at STP, will be evolved when 63 g of ammonium dichromate is decomposed? (H = 1, N = 14, O = 16, Cr = 52)

(N H_{4})_{2} C r_{2} O_{7} h e a t \to N_{2} (g) + 4 H_{2} O (g) + C r_{2} O_{3}

The volume in litres of dm

^{3}

N_{2}

evolved when 63 gm of ammonium dichromate is heated at STP is____________. (Please give your answer in the form of nearest integer)

(N H_{4})_{2} C r_{2} O_{7} \to C r_{2} O_{3} + 4 H_{2} O + N_{2}

The mass of

C r_{2} O_{3}

formed by the decomposition of 63 gm of ammonium dichromate :

[N = 14, H = 1, O = 16, C r = 52]

Q. Ammonium dichromate on heating decompose to form nitrogen and chromium ( lll) oxide and water what will be the loss in mass if 31.5 gram of Ammonium dichromate is heated at 100°C
(chromium=52,nitrogen=14,hydrogen=1,oxygen =16)
(NH4 )²Cr2O7 gives N2+Cr2O3+4H2O

Solid ammonium dichromate decomposes as the equationr:

$(N H_{4} 4)_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O$
If 63 g of sammonium dichromate decomposes, calculate:
(a) the quantity in moles of $(N H_{4})_{2} C r_{2} O_{7}$
(b) the qunatity in moles of nitrogen formed
(c) the volume of $N_{2}$ evolved at STP

(d) the loss of mass
(e) the mass of chromium (iii) oxide formed at the same time.

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(NH4)2Cr2O7→Cr2O3+4H2O+N2Calculate the volume of nitrogen at STP, evolved when 63 g of ammonium dichromate is heated.

The correct option is A 5.6 litre(NH4)2Cr2O7→Cr2O3+4H2O+N2One mole of ammonium dichromate will release one mole of nitrogen gas.252 gm of ammonium dichromate will release 22.4 litre nitrogen gas. So, 63 gm will give 22.4×63252=22.44=5.6 L N2.

$(N H_{4})_{2} C r_{2} O_{7} \to C r_{2} O_{3} + 4 H_{2} O + N_{2}$
Calculate the volume of nitrogen at STP, evolved when 63 g of ammonium dichromate is heated.