AP Chemistry question.. please help!!!!?

Question

a .500-g mixture of Cu2O and CuO contains .425 g Cu. What is the mass of CuO in the mixture???

also A 3.41g smaple of a metallic element, M, reacts completley with .0158 mol of a gas, X2, to form 4.52 g MX. What are the identities of M and X

Answer 1

mass of CuO = m, mass of Cu2O = 0.5 - m

MW (CuO) = 79.55 g/mol, MW (Cu2O) = 143.1 g/mol

(1 mol of Cu/mol CuO)*(m/79.55) = (m/79.55) mols Cu

(2 mol Cu/mol Cu2O) * [(0.5-m)/143.1] = 2*(0.5-m)/143.1 mols Cu

There are a total of 0.425 g Cu in mixture, or 0.425/63.55 mols Cu. Add up the moles of Cu in each of the substances, set equal to the total, and solve for m:

m/79.55 + 2*(0.5-m)/143.1 = 0.425/63.55

143.1m + 2*79.55*(0.5-m) = 79.55*143.1*(0.425/63.55)

-16m + 79.55 = 76.13

m ~ 0.214 g CuO
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The following equation governs the metal's rxn with the gas:

2M + X2 ------> 2 MX

0.0316 mols 0.0158 mols 0.0316 mols

The mass of metal M that reacted was stated to be 3.41 g and we also know the amount of moles that correspond to that mass, 0.0316. That's all we need to calculate M's MW:

MW (M) = 3.41 g/0.0316 mols ~ 107.91 g/mol

The metal M is Ag (silver).

From the equation, we know that 0.0316 mols of MX gas was produced and that the resulting mass was 4.52 g. Combining these facts with the MW of Ag will give us the MW of X and its corresponding element.....

(MW[X] + 107.91) * 0.0316.= 4.52

MW(X) ~ 35.13

The only possible element that X could be is Cl (chlorine).

Answer 2

.500-.425= .075g of Oxygen

.425*(1/3)= .1416g each of Cu (since there's 3)

.075/2= .0375g each O (since there's 2)

so... .1416+.0375= .1791g CuO


but I could be wrong.