What mass of N2H4 can be oxidised to N2 by 24.0 g K2CrO4 , which is reduced to Cr ( OH )4- (Given: Molar mass of K 2 CrO 4=194.2 )

Question

What mass of N2H4 can be oxidised to N2 by 24.0 g K2CrO4 , which is reduced to Cr ( OH )4- (Given: Molar mass of K 2 CrO 4=194.2 ) (A) 9.97 g (B) 2.97 g (C) 3.97 g (D) 4.97 g

Tardigrade Admin · Accepted Answer

N22 - arrow N20+ 4e C r6 + + 3 e arrow C r3 + Meq. of N2H4 = Meq. of K2C r O4 (w /32 / 4)× 1000 =(24/194.2 / 3)× 1000 (E . w t = (M w t ./v a l e n c e f a c t o r) ) textw textN text2 textH text4 text= 2 text.97 textg

Q. What mass of $N_{2} H_{4}$ can be oxidised to $N_{2}$ by 24.0 g $K_{2} C r O_{4}$ , which is reduced to $C r (O H)_{4}^{-}$ (Given: Molar mass of $K_{2} C r O_{4} = 194.2$ )

9.97 g

2.97 g

3.97 g

4.97 g

$N_{2}^{2 -} \to N_{2}^{0} + 4 e$

$C r^{6 +} + 3 e \to C r^{3 +}$

Meq. of $N_{2} H_{4}$ = Meq. of $K_{2} C r O_{4}$

$\frac{w}{32/4} \times 1000 = \frac{24}{194.2/3} \times 1000$ $(E . wt = \frac{M wt .}{v a l e n ce f a c t or})$

$w_{N_{2} H_{4}} = 2 .97 g$

Q. What mass of N2​H4​ can be oxidised to N2​ by 24.0 g K2​CrO4​ , which is reduced to Cr(OH)4−​ (Given: Molar mass of K2​CrO4​=194.2 )