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Chemistry

From the equation : (NH4)2 Cr2O7 ⟶ Cr2O3 + 4H2O + N2 Calculate :

(i) the vol. of nitrogen at STP, evolved when 63 g. of ammonium dichromate is heated.

(ii) the mass of Cr2O3 formed at the same time.

[N = 14, H = 1, Cr = 52, O = 16]

Stoichiometry

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Answer

(NH4)2Cr2O7→ΔCr2O3+4H2O+ N22[14+4(1)]2(52)1 mol.+2(52)+7(16)+3(16)vol.=36+104+112=104+48=22.4 lit=252 g=152 g\begin{matrix} (\text{NH}4)2\text{Cr}2\text{O}7 & \xrightarrow{\Delta} & \text{Cr}2\text{O}3 & + & 4\text{H}2\text{O} & + \text{ N}2 \ 2[14 + 4(1)] & & 2(52) & & & 1 \text{ mol.} \ + 2(52) + 7(16) & & + 3(16) & & & \text{vol.} \ = 36 + 104 + 112 & & = 104 + 48 & & & = 22.4 \text{ lit} \ = 252 \text{ g} & & = 152 \text{ g} \ \end{matrix}

252 g of ammonium dichromate gives 22.4 lit of nitrogen

∴ 63 g of ammonium dichromate will give 22.4252\dfrac{22.4}{252} x 63 = 5.6 lit of nitrogen.

Hence, 5.6 lit of nitrogen is evolved.

(ii) 252 g of ammonium dichromate gives 152 g Cr2O3

∴ 63 g of ammonium dichromate will give 152252\dfrac{152}{252} x 63 = 38 g

Hence, 38 g of Cr2O3 is formed.

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