CuO--> Cu= 1 Cu/ total> 100= 64/80= 80%Cu/20%O
O= 1*16
sample 2:
Cu2O--> 2*64 128 128/144> 100= 89%Cu/ 11%O
1*16 16
144 g/mol
Step by step process. . .
Write a balance chemical equation and identify the knowns and the unknowns.
25.0 50.0g ---> ?g
P'4(s) + 50'2(g) ---> P'4O'1O(s)
Determine the number of moles of the reactants by multiplying each mass by the conversation factor that relates moles and mass, the inverse of molar mass.
25.0 gP'4 * 1 mol P'4/123.9 g P'4 = .202 mol P'4
50 g O'2 * 1 mol O'2/32 g O'2= 1.56 mol O'2
Calculate the actual ratio of available moles of 0'2 and available moles of P'4.
1.56 mol O'2/.0202 mol P'4 = 7.72 mol O'2/ 1 mol P'4
Determine the mole ration of the two reactants fomr the balanced chemical equation.
5 mol O'2/ 1 mol P'4
Because 7.72 mol O'2 is available but only 5 mol is needed to react with 1 mol P'4, O'2 is in excess and P'4 is the limiting reactant. Use the moles of P'4 to determine the moles of P'4O'10 that will be produced.
Multiply the number of moles of P'4 by the mole ratio of P'4O'10 (the unknown) to P'4 (the unknown)
.0202 mol P'4 * 1 mol P'4O'10/1 mol P'4 = .202 mol P'4 O'10
and then i will add the yellow paper from the scanner in class. . .
