Complete and balance the reaction of

A) (CH3)2NH(aq)+CH3COOH(aq)→

B) CH3CH2NH2(aq)+HBr(aq)→

C) CH3NH2(aq)+HCOOH(aq)→

A) To balance the reaction (CH3)2NH(aq) + CH3COOH(aq), we need to make sure that the number of atoms on each side of the equation is the same.

The balanced equation is:
(CH3)2NH(aq) + CH3COOH(aq) → (CH3)2NCH3COO(aq)

B) To balance the reaction CH3CH2NH2(aq) + HBr(aq), we need to make sure that the number of atoms on each side of the equation is the same.

The balanced equation is:
CH3CH2NH2(aq) + HBr(aq) → CH3CH2NH3Br(aq)

C) To balance the reaction CH3NH2(aq) + HCOOH(aq), we need to make sure that the number of atoms on each side of the equation is the same.

The balanced equation is:
CH3NH2(aq) + HCOOH(aq) → CH3NH3+COO-(aq)

To balance chemical equations, we need to ensure that the same number of atoms of each element are present on both sides of the equation. Additionally, we want to make sure the charges are balanced as well.

A) (CH3)2NH(aq) + CH3COOH(aq) →

First, let's start by balancing the carbon and hydrogen atoms:

(CH3)2NH: 2 carbon atoms and 8 hydrogen atoms
CH3COOH: 2 carbon atoms, 4 hydrogen atoms, and 2 oxygen atoms

We can balance the carbon atoms by placing a coefficient of 2 in front of CH3COOH:

(CH3)2NH(aq) + 2CH3COOH(aq) →

Now, let's balance the hydrogen atoms:

(CH3)2NH: 2 carbon atoms and 8 hydrogen atoms
2CH3COOH: 2 carbon atoms, 8 hydrogen atoms, and 4 oxygen atoms

Since the number of hydrogen atoms are already balanced, we can move on to balancing the oxygen atoms:

(CH3)2NH: 2 carbon atoms and 8 hydrogen atoms
2CH3COOH: 2 carbon atoms, 8 hydrogen atoms, and 4 oxygen atoms

The oxygen atoms are already balanced, so the balanced equation is:

(CH3)2NH(aq) + 2CH3COOH(aq) → (CH3)2NCH3COOH(aq)

B) CH3CH2NH2(aq) + HBr(aq) →

Let's balance the carbon, hydrogen, and bromine atoms:

CH3CH2NH2: 3 carbon atoms, 8 hydrogen atoms, and 1 nitrogen atom
HBr: 1 hydrogen atom and 1 bromine atom

We can balance the bromine atoms by placing a coefficient of 2 in front of HBr:

CH3CH2NH2(aq) + 2HBr(aq) →

Now, let's balance the hydrogen atoms and nitrogen atom:

CH3CH2NH2: 3 carbon atoms, 8 hydrogen atoms, and 1 nitrogen atom
2HBr: 2 hydrogen atoms and 2 bromine atoms

Since the number of hydrogen atoms and nitrogen atoms are already balanced, the balanced equation is:

CH3CH2NH2(aq) + 2HBr(aq) → CH3CH2NH3Br(aq)

C) CH3NH2(aq) + HCOOH(aq) →

Let's balance the carbon, hydrogen, and oxygen atoms:

CH3NH2: 1 carbon atom, 5 hydrogen atoms, and 1 nitrogen atom
HCOOH: 1 carbon atom, 2 hydrogen atoms, and 2 oxygen atoms

We can balance the carbon atoms by placing a coefficient of 2 in front of CH3NH2:

2CH3NH2(aq) + HCOOH(aq) →

Next, let's balance the hydrogen atoms and oxygen atoms:

2CH3NH2: 2 carbon atoms, 10 hydrogen atoms, and 2 nitrogen atoms
HCOOH: 1 carbon atom, 2 hydrogen atoms, and 2 oxygen atoms

To balance the hydrogen atoms, we need to place a coefficient of 5 in front of HCOOH:

2CH3NH2(aq) + 5HCOOH(aq) →

Now, let's check if the oxygen atoms are balanced:

2CH3NH2: 2 carbon atoms, 10 hydrogen atoms, and 2 nitrogen atoms
5HCOOH: 5 carbon atoms, 10 hydrogen atoms, and 10 oxygen atoms

The oxygen atoms are already balanced, so the balanced equation is:

2CH3NH2(aq) + 5HCOOH(aq) → 2CH3NHCOOH(aq) + 3H2O(aq)

This means that for reaction A, (CH3)2NH(aq) + CH3COOH(aq) → (CH3)2NCH3COOH(aq)
For reaction B, CH3CH2NH2(aq) + 2HBr(aq) → CH3CH2NH3Br(aq)
For reaction C, 2CH3NH2(aq) + 5HCOOH(aq) → 2CH3NHCOOH(aq) + 3H2O(aq)

I have consulted with a friend who is an organic chemist and he tells me all of these are acid/base reactions. Hence for part B

CH3CH2NH2 + HBr ==> CH3CH2NH3^+ + Br^- or you may want to write that as CH3CH2NH3Br. That is the salt but is ionic and ionizes as I have it above.