Q1 ) In a laboratory experiemnt you completed this semester, the percent sulfate

Question

Q1 ) In a laboratory experiemnt you completed this semester, the percent sulfate in an unknown sample was determined using a gravimetric procedure. In a similar experiment the percent copper can also be determined by precipitation of copper(II) carbonate. A 0.1246 g sample containing an unknown amount of copper was dissolved in water and treated with an excess of H2CO3. The resulting dried precipitate weighed 0.0816 g. Determine the percent of copper in the original sample.

66.3 %

65.5 %

33.7 %

48.6 %

51.4 %

Q2) If 258 mL of a 0.500 M Pb(NO3)2 solution is mixed with 500 mL of a 0.312 M NaI solution, what is the identity and mass of the precipitate that is formed?

Precipitate = PbI2, mass = 59.5 g

Precipitate = PbI2, mass = 36.0 g

Precipitate = PbI2, mass = 18.0 g

Precipitate = NaNO3, mass = 5.48 g

Precipitate = NaNO3, mass = 3.32 g

66.3 %

65.5 %

33.7 %

48.6 %

51.4 %

Explanation / Answer

The equation determined should read Cu + H2CO3 ----> CuCO3 + H2   ....... But discount the H2, you dont need it at all. The given is the mass of the precipitate, which is .0816g CuCO3

To solve: .0816g CuCO3 * (1mol CuCO3/123.555g[a.k.a. molar mass]) = 0.000660435 mol CuCo3. Multpily this by the mole ratio of CuCo3 to Cu to get moles of Cu, in this case 1:1. So no change in the moles. Then multiply .000660435*63.546gCu = .04262g Cu. Finally divide .04262g/.1246g= 33.7 %

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