For the reaction: $2X + 3Y \rightarrow 3Z$, the combination of $2.00$ moles of $X$ with $2.00$ moles of $Y$ produces $1.75 $ moles of $Z$. What is the percent yield of this reaction?
  
$\textbf{(A)}\hspace{.05in}43.8\%\qquad\textbf{(B)}\hspace{.05in}58.3\%\qquad\textbf{(C)}\hspace{.05in}66.7\%\qquad\textbf{(D)}\hspace{.05in}87.5\%\qquad $
1. **Determine the limiting reagent:**
   - The balanced chemical equation is: 
     \[
     2X + 3Y \rightarrow 3Z
     \]
   - We start with 2.00 moles of \(X\) and 2.00 moles of \(Y\).
   - The stoichiometric ratio from the balanced equation is 2 moles of \(X\) to 3 moles of \(Y\).
   - Calculate the moles of \(Y\) required to react with 2.00 moles of \(X\):
     \[
     \text{Moles of } Y = 2.00 \text{ moles } X \times \frac{3 \text{ moles } Y}{2 \text{ moles } X} = 3.00 \text{ moles } Y
     \]
   - Since we only have 2.00 moles of \(Y\), \(Y\) is the limiting reagent.

2. **Calculate the theoretical yield of \(Z\):**
   - Using the limiting reagent \(Y\), calculate the moles of \(Z\) produced:
     \[
     \text{Moles of } Z = 2.00 \text{ moles } Y \times \frac{3 \text{ moles } Z}{3 \text{ moles } Y} = 2.00 \text{ moles } Z
     \]

3. **Calculate the percent yield:**
   - The actual yield of \(Z\) is given as 1.75 moles.
   - The percent yield is calculated using the formula:
     \[
     \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%
     \]
   - Substitute the values:
     \[
     \text{Percent Yield} = \left( \frac{1.75 \text{ moles } Z}{2.00 \text{ moles } Z} \right) \times 100\% = 87.5\%
     \]

Conclusion:
\[
\boxed{87.5\%}
\]