If You are given 14 moles of C2H5OH(Ethanol) and 13 moles of O2( Oxygen), how many moles of the limiting reactant is produce? C2H5OH + O2 -> CH3COOH +
H2O

Explanation:

1 ) first u have to balance the equation but in this case it is already balanced.

2) checking for the limiting reactant:

13 mol O2 × (1mol H2O / 1mol of O2) = 13 mol of H2O

14 mol of ethanol × (1mol H2O/ 1molof ethanol) = 14 mol of H2O

the less amount of moles produced from the O2 so the limiting reactant is oxygen

3) calculating the products moles

13 mol O2 × (1mol H2O / 1mol of O2) = 13 mol

13 mol O2 × (1mol Ch3Cooh / 1mol O2) = 13 mol

note: the ratios obtained from the balanced equation

4) the excess reagent is the ethanol 1 mol will be remain

uranium-238 i believe is your answer

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it