Calculate the mass in 3 mols of NH4OH and calculate the mass in 4.2 moles of Ca(NO3)2.

The formula indicates 1 N atom, 5 H atoms (4 from $\text{NH}_4$ and 1 from OH), and 1 O atom.
$$
\text{Molar mass of NH}_4\text{OH} = (1 \times \text{N}) + (5 \times \text{H}) + (1 \times \text{O})
$$
$$
\text{Molar mass} = (1 \times 14.01 \, \text{g/mol}) + (5 \times 1.01 \, \text{g/mol}) + (1 \times 16.00 \, \text{g/mol})
$$
$$
\text{Molar mass} = 14.01 \, \text{g/mol} + 5.05 \, \text{g/mol} + 16.00 \, \text{g/mol}
$$
$$
\text{Molar mass} = 35.06 \, \text{g/mol}
$$

The formula for mass ($m$) is:
$$
m = \text{moles (n)} \times \text{molar mass (M)}
$$
$$
m = 3 \, \text{mol} \times 35.06 \, \text{g/mol}
$$
$$
m = 105.18 \, \text{g}
$$
The mass in 3 moles of $\text{NH}_4\text{OH}$ is $\boxed{\text{105.18 g}}$.

The formula indicates 1 Ca atom, 2 N atoms (from $2 \times \text{NO}_3$), and 6 O atoms (from $2 \times 3 \text{ O}$).
$$
\text{Molar mass of Ca(NO}_3)_2 = (1 \times \text{Ca}) + (2 \times \text{N}) + (6 \times \text{O})
$$
$$
\text{Molar mass} = (1 \times 40.08 \, \text{g/mol}) + (2 \times 14.01 \, \text{g/mol}) + (6 \times 16.00 \, \text{g/mol})
$$
$$
\text{Molar mass} = 40.08 \, \text{g/mol} + 28.02 \, \text{g/mol} + 96.00 \, \text{g/mol}
$$
$$
\text{Molar mass} = 164.10 \, \text{g/mol}
$$

$$
m = \text{moles (n)} \times \text{molar mass (M)}
$$
$$
m = 4.2 \, \text{mol} \times 164.10 \, \text{g/mol}
$$
$$
m = 689.22 \, \text{g}
$$
The mass in 4.2 moles of $\text{Ca(NO}_3)_2$ is $\boxed{\text{689.22 g}}$.