Complete and balance the reactions
A) (CH3)2NH(aq)+CH3COOH(aq)→
B) CH3CH2NH2(aq)+HBr(aq)→
C) CH3NH2(aq)+HCOOH(aq)→
All of these are acid/base reactions.
B.
CH3CH2NH2 + HBr ==> CH3CH2NH3^+ + Br^- or as a compound CH3CH2NH3Br
A) (CH3)2NH(aq)+CH3COOH(aq)→(CH3)2NH2+(aq)+CH3COO^-(aq)
B) CH3CH2NH2(aq)+HBr(aq)→CH3CH2NH3+(aq)+Br^-(aq)
C) CH3NH2(aq)+HCOOH(aq)→CH3NH3+(aq)+HCOO^-(aq)
To complete and balance these reactions, we need to first identify the reactants and products.
A) (CH3)2NH(aq) + CH3COOH(aq) → ?
In this case, (CH3)2NH is a compound known as dimethylamine and CH3COOH is acetic acid. To complete the reaction, we would expect the formation of a salt. The reaction can be represented as follows:
(CH3)2NH(aq) + CH3COOH(aq) → (CH3)2NH2+CH3COO-(aq)
B) CH3CH2NH2(aq) + HBr(aq) → ?
In this case, CH3CH2NH2 is ethylamine and HBr is hydrobromic acid. When these two react, we would expect the formation of a salt as well. The reaction can be represented as follows:
CH3CH2NH2(aq) + HBr(aq) → CH3CH2NH3+Br-(aq)
C) CH3NH2(aq) + HCOOH(aq) → ?
In this case, CH3NH2 is methylamine and HCOOH is formic acid. When these two react, we would also expect the formation of a salt. The reaction can be represented as follows:
CH3NH2(aq) + HCOOH(aq) → CH3NH3+HCOO-(aq)
To balance the equations, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In these cases, since we have already assigned charges to the ions, the equations are already balanced.
Note: It's important to consider the state of the compounds (aq for aqueous solution) and the charges on the ions when balancing equations.