Question

COMPLETE COMBUSTION of C10H10O2 to CO2 and H2O. You have 204.3566 grams of C10H10O2 and 735.972 grams of O2. What is the limiting reagent? (I arrived at the answer of C10H10O2) How many grams of CO2 are generated? How many grams of H2O are generated?

Answer 1

Limiting substance:

Explanation:

`C_10H_10O_2 + 11.5O_2 -> 10CO_2 + 5H_2O`

is the chemical reaction (for full combustion)

MW (fuel) = 162 grams

How many moles of fuel= `204.3566/162 = 1.26` moles
How many moles of oxygen gas do we have
`=735.972/32 = 22.99` moles

1 mole od fuel needs 11.5 moles of oxygen gas but you have only 1.26 moles of fuel. It means you need 14.49 moles of oxygen is required to fully consume the available fuel. Therefore your limiting matter is fuel.

Carbondioxide generated `= 12.6` moles of carbondioxide is formed. It is `12.6times44 = 554.4` grams of carbondioxide.

Water vapor produces `=1.26times5 = 6.3` moles. It is 113.4 grams of water vapor.