Step-by-step explanation:
The balanced equation for the reaction is given as;
2CH3CHO + O2 → 2CH3COOH
If 20.0 g CH3CHO and 10.0 g O2 were put into a reaction vessel, (a)
how many grams of acetic acid will be produced?
First thing's first, we have to find he limiting reactant. This is done by comparing the number of moles of the reactants.
From the equation of the reaction;
2 mol of CH3CHO reacts with 1 mol of O2
From the masses given;
Number of moles = Mass / Molar mass
CH3CHO;
Number of moles = 20 / 44.0526 = 0.454 mol
O2;
Number of moles = 10 / 32 = 0.3125 mol
The limiting reactant is CH3CHO because O2 would be in excess.
Back to the question;
2 mol of CH3CHO produces 2 mol of CH3COOH
0.454 mol would produce x
Solving for x;
x = 0.454 * 2 / 2 = 0.454 mol
Converting to mass;
Mass = number of moles* Molar mass
Mass = 0.454 mol * 60.052 g/mol = 27.26 grams
(b) how many grams of the excess reactant remain after the reaction is
complete
The excess reactant is O2
Number of moles left = Initial Number of moles - Number of moles that reacted
Number of moles left = 0.3125 mol - (0.454 mol / 2)
Number of moles left = 0.0855 mol
Converting to mass;
Mass = 0.0855 mol * 32 g/mol = 2.736 grams