What is the value of enthalpy change for the following reaction
$$\ce{X + 2Y -> 2Z}$$
given
\begin{align} \ce{W + X &-> 2Y} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-200 kcal} \tag{1}\\ \ce{2W + 3X &-> 2Z + 2Y} &\qquad & \Delta_\mathrm{r}H_2 = \pu{-150 kcal} \tag{2} \end{align}
I have the answers, and the correct answer is $\pu{250 kcal}$, so I'm assuming you do $400-150$, but not sure how you would go about it.
$$\ce{X + 2Y -> 2Z}$$
\begin{align} \ce{W + X &-> 2Y} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-200 kcal} \tag{1}\\ \ce{2W + 3X &-> 2Z + 2Y} &\qquad & \Delta_\mathrm{r}H_2 = \pu{-150 kcal} \tag{2} \end{align}
$(-1) \times 2$ + $(2)$ gives:
$\ce{4Y +2W + 3X->2W + 2X + 2Z + 2Y} \equiv \ce{2Y +X -> 3Z}$
Using Hess's Law, the enthalpy change for the required reaction is given by $-2\times \Delta_rH_1+\Delta_rH_2= 400-150 = \pu{250 kcal}$
