0.260 mol of octane is allowed to react with 0.900 mol of oxygen a) How many mol

Question

0.260 mol of octane is allowed to react with 0.900 mol of oxygen

a) How many moles of water are produced in this reaction?

b) After the reaction, how much octane is left?

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)-->2AlCl3(s). You are given 18.0 g of aluminum and 23.0 g of chlorine gas

c) If you had excess chlorine, how many moles of aluminum chloride could be produced from 18.0 g of aluminum?

d) If you had excess aluminum, how many moles of aluminum chloride could be produced from 23.0 g of chlorine gas, Cl2?

Explanation / Answer

Part A

a) Chemical equation,

2C8H18 + 25O2 ---> 16CO2 + 18H2O

moles of octane = 0.260 mols

moles of O2 = 0.900 mols

If all of octane has reacted we would need = 0.26 x 25/2 = 3.25 moles of O2

If all of O2 has reacted we would require = 0.9 x 2/25 = 0.072 moles of octane

Since the moles of octane is greater then what is required, O2 is the limiting reagent

moles of H2O formed = 0.9 x 18/25 = 0.648 mols

b) After the reaction the moles of octane left = 0.260 - 0.072 = 0.188 mols

Part B.

2Al(s) + 3Cl2(g) ---> 2AlCl3(s)

moles of Al = 18/26.9815 = 0.667 mols

moles of Cl2 = 23/70.906 = 0.324 mols

c) moles of AlCl3 produced from Al = 0.667 mols

d) moles of AlCl3 produced from Cl2 gas = 0.324 x 2/3 = 0.216 mols

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