Combustion analysis of 2.400 g of an unknown compound containing carbon, hydrogen, and oxygen produced 4.171 g of CO2and 2.268 g of H2O. What is the empirical formula of the compound

Explanation:

Hello,

In this case, for the given combustion process, we can compute the present moles of carbon atoms by considering that the only product containing carbon is CO2, thus, we have:

Next, we compute the moles of hydrogen that is present at the water only as a product:

Then, we compute the moles of oxygen by considering the oxygen present in both the CO2 and H2O and the initial sample:

Now, we divide the moles of C, H and O by the moles of O since they are lower amount, in order to compute the subscript in the empirical formula:

But the smallest whole number is:

Regards.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2