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A 21.5 g sample of nickel was treated with excess silver nitrate solution to produce silver metal and nickel(II) nitrate. The reaction was stopped before all the nickel reacted, and 36.5 g of solid metal (nickel and silver) is present. Calculate the mass of solid silver metal present in grams.

Respuesta :

The mass of nickel is 36.5g metal.

Ni+2AgNO3 → Ni(NO3)2+2Ag

21.5g Ni / 58.69g = 0.366mol Ni

Ag = 107.87

N i= 58.69

But, 2molAg per 1 mol Ni, so, change in mass is more like...

Twice as many moles Ag as Ni, and then the mass relationship is 107.87g Ag / 58.69, but due to the 2Ag: 1Ni mol ratio, the change in mass is like:

107.87(2) / 58.69 = 3.68 times increase

So, the final mass of Ag is 3.68x, and the change in mass of Ni is x

36.5=(21.5-x)+(3.68x)

Solve for x

x=5.597

3.68x=g Ag

3.68(5.597)=20.59

21.5 - x = g Ni after reaction

21.5 - 5.597 = 15.903g Ni

20.59g Ag + 15.903g Ni = 36.5g metal

The mass of nickel is 36.5g metal.

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