Question

How many grams of oxygen is required for complete combustion of 12g of aluminum?

Answer 1

Answer:

9.6kg

Explanation:

Your strat should be to 1) convert 2.8 kg ethylene (Et) to moles, 2) convert moles ethylene to moles oxygen gas, finally, 3) convert moles oxygen gas to mass oxygen gas.

1) number moles, n = mass Et/(molar weight Et) = 2800 g/ (28 g/mol).

There are 100 moles of Et.

2) The balanced equation is C2H4 + 3 O2 -> 2 CO2 + 2 H2O. This tells us that we need 3 oxygen molecules for every Et molecule. So we need 300 moles of Et.

3) To find the mass of 300 moles oxygen gas, use mass = n.(molar weight O2). Mass O2 = 300 moles x (32.0 g/mol) = 9600 g O2

Answer = 9.6 kg