#1 A mixture of calcium sulfate (CaSO 4 ) and calcium uranate (CaUO 4 ) contains

Question

#1 A mixture of calcium sulfate (CaSO4) and calcium uranate (CaUO4) contains 10% sulfur by mass.

What i the mass percentage of uranium in the mixture?                

#2 A mixture containing 50 g of carbon (C) and 50 g of lime (CaO) is reacted at high temperature to produce calcium carbide (CaC2) according to the reaction

CaO(s) + 3 C(s) ? CaC2(s) + CO(g)

Assuming that the reaction goes to completion and that one of the reactants is completely consumed, what is the weight percent of unreacted reagent in the solid residue produced?

A) 90.7            B) 46.6            C) 15.6            D) 85.4            E) 23.8

Explanation / Answer

1.) Molar mass of S = 32.065 g/mole
Molar mass of CaSO4 = 136.1406 g/mole
Molar mass of U = 238.0289 g/mole
Molar mass of CaUO4 = 342.1045 g/mole
Start by assuming there is 100 grams of the mixture, total. This means there are 10 grams of S
10g S*(136.14/32.065) = 42.46 grams CaSO4
100g - 42.46g = 57.54 grams CaUO4
57.54g CaUO4*(238.0289/342.1045) = 40 grams U
40.0% U by mass

2.) 50g C*(1 mol/12.0107g) = 4.16 mol C
50g CaO*(1mol/56.077g) = 0.892 mol CaO
CaO is limiting reactant
3*0.892mol = 2.67 mol C reacts
4.16 - 0.892 mol = 1.89 mol C unreacted = 17.9 gC
0.892 mol CaC2 formed = 57.2 g CaC2
17.9/(57.2+17.9) = 0.238 = 23.8%
23.8% (E)

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