Write the reaction and given data

$X + Y + 3Z \rightarrow XYZ_3$

Given initial amounts:

• $X = 1$ mol
• $Y = 1$ mol
• $Z = 0.05$ mol

Find the limiting reagent

From the stoichiometry:

$3 \text{ mol of } Z \text{ produce } 1 \text{ mol of } XYZ_3$

Hence, with $0.05$ mol of $Z$:

$\text{moles of } XYZ_3 = \frac{0.05}{3} = 0.01667 \text{ mol}$

Since $X$ and $Y$ are present in excess (1 mol each), $Z$ is the limiting reagent.

Calculate molar mass of $XYZ_3$

Atomic masses are:

• $X = 10$
• $Y = 20$
• $Z = 30$

Therefore,

$M(XYZ_3) = 10 + 20 + 3(30) = 120 \text{ g mol}^{-1}$

Calculate mass of product formed

$\text{mass} = \text{moles} \times \text{molar mass}$

$= \frac{0.05}{3} \times 120 = 2.0 \text{ g}$

Final answer

The yield of $XYZ_3$ is

$\boxed{2 \text{ g}}$

Nearest integer = $\boxed{2}$