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

The mass of grams of $C r_{2} O_{3}$ formed during the reaction is________. [Atomic masses: H=1, Cr=52, N=14]

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Solution

Gram molecular mass of $(N H_{4})_{2} C r_{2} O_{7}$ = $252 g$ .

Gram molecular mass of $N_{2}$ = $28 g .$

Gram molecular mass of $C r_{2} O_{3}$ = 1 $52 g .$

If $252 g$ of $(N H_{4})_{2} C r_{2} O_{7}$ release $152 g$ of $C r_{2} O_{3}$

then, $63 g$ of $(N H_{4})_{2} C r_{2} O_{7}$ will release = $= \frac{152 \times 63}{252} = 37.99 g$ $\approx 38$ .

Hence, the correct answer is $38$ .

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(NH4)2Cr2O7heat→N2(g)+4H2O(g)+Cr2O3The mass of grams of Cr2O3 formed during the reaction is________. [Atomic masses: H=1, Cr=52, N=14]

Gram molecular mass of (NH4)2Cr2O7 = 252g.Gram molecular mass of N2 = 28g.Gram molecular mass of Cr2O3 = 152g.If 252g of (NH4)2Cr2O7 release 152g of Cr2O3then, 63g of (NH4)2Cr2O7 will release = =152×63252=37.99g ≈38. Hence, the correct answer is 38.

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

The mass of grams of $C r_{2} O_{3}$ formed during the reaction is________. [Atomic masses: H=1, Cr=52, N=14]