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

answer; protein;

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6