physical chemistry - Using enthalpies of formation and combustion to find formation enthalpy of Acetylene (Hess' Law)

I understand this is a simple question (I’m in high school), but for some reason the numbers just are not working for me.

Given: $$\Delta H^\circ_\text{c} = -1301\ \mathrm{kJ}\ \text{for}\ \ce{C2H2(g)}$$ $$\Delta H^\circ_\text{f} = -394\ \mathrm{kJ}\ \text{for}\ \ce{CO2(g)}$$ $$\Delta H^\circ_\text{f} = -242\ \mathrm{kJ}\ \text{for}\ \ce{H2O(g)}$$ Calculate $\Delta H_\text{f}$ for $\ce{C2H2(g)}$.

My process: $$(1) : \ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(g) }\quad \Delta H^\circ_\text{r} = -1301\ \mathrm{kJ}$$ $$(2) : \ce{C(s) + O2(g) -> CO2(g)} \quad \Delta H^\circ_\text{r} = -394\ \mathrm{kJ}$$ $$(3) : \ce{H2(g) + 1/2O2(g) -> H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$

$$\text{Target:}\ \ce{2C(s) + H2(g) -> C2H2(g)}$$

After some experimentation I arrive at:

$$-1\times(1) + 2\times(2) + (3) = \text{Target}$$

As shown through: $$-1\times(1){:}\ \ce{2CO2(g) + H2O(g) -> C2H2(g) + 5/2 O2(g)} \quad \Delta H^\circ_\text{r} = 1301\ \mathrm{kJ}$$ $$2\times(2){:}\ \ce{2C(s) + 2O2(g) -> 2CO2(g)} \quad \Delta H^\circ_\text{r} = -788\ \mathrm{kJ}$$ $$(3){:}\ \ce{H2(g) + 1/2O2(g) -> H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$ $$\text{Sum:}\ \ce{2C(s) + H2(g) -> C2H2(g)} \quad \Delta H^\circ_\text{f} = 271\ \mathrm{kJ}$$

Now as far as I know, all the givens are accurate empirical values. But the empirical value for $\Delta H^\circ_\text{f}$ for $\ce{C2H2(g)}$ is $227\ \mathrm{kJ}$.

What’s even weirder is that this disparity $(271-227=44)$ is equal to the difference between standard enthalpies of formation for water in gas and liquid states $(-242 - -286 = 44)$.

Have I made an error somewhere, or is water formed in the liquid state during the combustion of acetylene?