Question

2C₄H₉OH + 8 O₂  →  8 CO  + 10 H₂O  DH = -2710 kJ

If you have 5.77 g of the organic fuel (MW = 74.12 g/mol)  and 10.77 g of oxygen gas, how many grams of carbon monoxide will you produce?

A) Write the reaction

B) put a square around the compound that will be the precipitate

C) how many moles of each reactant do you have?

D) How many grams of the precipitate did you create

Extra Credit:

How many moles of excess reactant do you have left at the end of the reaction?

Answer 1

A)

The balanced chemical equation would be,

2C4H9OH + 8 O2 ====> 8 CO + 10 H2O

B)

There is no formation of precipitate in the reaction process as we can see from the chemical equation.

C)

Mol of organic fuel = mass / (molar mass x stoichiometric coefficient)

= 5.77 g / (74.1gmol-1 x 2)

= 0.039

Similarly, mol of O2 = 10.77 g / (32 g mol-1 x 8)

= 0.0421 mol

Since, mol of organic fuel are the least it would be the limiting reagent in the reaction.

D) gram of CO produce =(mass of limiting reagent / molar mass of limiting reagent) x (stoichiometric coefficient of product / stoichiometric coefficient of Limiting reagent) x molar mass of product

= (5.77 g / 74.1 gmol-1) x (8 / 8) x 28.01 g mol-1

= 2.18 g

Mass of excess reagent consumed = (mass of limiting reagent / molar mass of limiting reagent) x (stoichiometric coefficient of excess reagent / stoichiometric coefficient of Limiting reagent) x molar mass of excess reagent

= (5.77 g / 74.01 gmol-1) x (8 / 2) x 32 gmol-1

= 9.98 g of O2 consumed

So, excess reactant left over = total mass - mass consumed

= (10.77 - 9.98) g

= 0.80 g