Question

7. One way to isolate metals from their ores is to react the metal oxide with carbon as shown in the
following reaction:
2MO(s) + C(s) -> 2M(s) + CO2(g)
If 34.49 g of a metal oxide reacted with excess carbon and 4.42 L of CO2 formed at 100°C and
1.50 atm, what is the identity of the metal?

Answer 1

Metal oxides react with carbon and are reduced to the metal. The metal in the reaction above is copper.

The equation of the reaction is;

2MO(s) + C(s) -> 2M(s) + CO2(g)

We are told in the question that carbon is in excess so the metal oxide is the limiting reactant.

Number of moles of metal oxide = 34.49 g/M + 16

From PV = nRT

P = 1.50 atm

V = 4.42 L

T =  100°C + 273  = 373 K

n =?

R = 0.082 atmLK-1mol-1

n = PV/RT

n = 1.50 atm ×  4.42 L/0.082 atmLK-1mol-1 × 373 K

n = 6.63/30.58

=0.217 moles

If 2 moles of metal oxide yields 1 mole of CO2

34.49 g/M + 16 yields 0.217 moles of CO2

34.49 g/M + 16 = 2 × 0.217

34.49 /M + 16 = 0.434

34.49 = 0.434(M + 16)

34.49 = 0.434M + 6.944

M = 34.49 -  6.944/0.434

M = 63.5

The metal is copper.

Learn more: https://brainly.com/question/22824409