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The addition of dimethylglycoxime, H2C4H6O2N2, to a solution containing nickel(II) ion gives rise to a precipitate: Ni2 2H2C4H6O2N2 Ni(H2C4H6O2N2)2 2H If 0.15 g nickel alloy is treated with dimethylglycoxime and .175 mg nickel dimethylglycoxime is collected. Determine the mass and percent of nickel in the alloy.

Sagot :

Solution :

The balanced equation is :

                              [tex]$Ni^{2+}+2H_2C_4H_6O_2N_2 \rightarrow Ni(H_2C_4H_6O_2N_2)_2+2H^+$[/tex]

Molar mass           56.7            116                        290.7

From the balanced equation,

2 mole

= 2 x 116 g of [tex]$H_2C_4H_6O_2N_2$[/tex] produces 1 mole = 290.7 g of nickel dimethylglycoxime

or 2 x 116 mg of [tex]$H_2C_4H_6O_2N_2$[/tex] produces 1 mole = 290.7 g of nickel dimethylglycoxime

0.175 mg of [tex]$H_2C_4H_6O_2N_2$[/tex] produces [tex]$\frac{0.175 \times 290.7}{2 \times 116}$[/tex] = 0.219 mg of nickel dimethylglycoxime

290.7 g of [tex]$Ni(H_2C_4H_6O_2N_2)_2$[/tex] contains 58.7 mg of Ni

0.219 mg of [tex]$Ni(H_2C_4H_6O_2N_2)_2$[/tex] contains [tex]$\frac{0.219 \times 58.7}{290.7} = 0.0443$[/tex]  mg of Ni

So mass of nickel, m = 0.0443 mg = [tex]$0.0443 \times 10^{-3}$[/tex] g

Percent of Nickel in the alloy = [tex]$\frac{\text{mass of nickel}}{\text{mass of alloy}} \times 100$[/tex]

                                                [tex]$=\frac{0.0443 \times 10^{-3}}{0.159}\times 100$[/tex]

                                               = 0.03%