if my reaction is 2Na + 2H2O -> 2NaOH +H2, and i have been given 1 gram of sodium, how would i measure the theoretical change in heat of the reaction, assuming the reactants are at room temperature?

Look in your text, probably in the appendix, for delta H0 tables. Those usually are given at 25 degrees C and that is about room temperature.
Heat of rxn = deltaH products - deltaH reactants. The numbers in the appendix are per mol (or kJ/mol), usually; therefore, multipy what you find for NaOH by 2 and what you find for water by 2, etc. Na and H2 will be zero.

To measure the theoretical change in heat of the reaction, you will need to refer to the delta H0 tables in your text, usually found in the appendix. These tables provide the standard enthalpy changes (∆H0) for various reactions at a given temperature, typically at 25 degrees Celsius, which is approximately room temperature.

The heat of reaction (∆H) can be calculated using the formula:

∆H rxn = ∆H products - ∆H reactants

Start by looking up the standard enthalpy change for the products and reactants involved in the reaction. In this case, you have:

Reactants:
2Na (sodium) - Since the delta H0 for elements in their standard states is generally considered to be zero, the value for sodium will be zero.

2H2O (water) - Find the delta H0 for water in the table. Since you have two moles of water in the reaction, you will need to multiply the delta H0 value by 2.

Products:
2NaOH (sodium hydroxide) - Look up the delta H0 for sodium hydroxide in the table and multiply it by 2 since you have two moles of sodium hydroxide in the reaction.

H2 (hydrogen gas) - Similar to the sodium, since hydrogen is an element in its standard state, its delta H0 will be zero.

Once you have all the delta H0 values, substitute them into the formula and calculate the theoretical change in heat of the reaction.

Remember to pay attention to the units and convert them if necessary. The values in the delta H0 table are often given as kilojoules per mole (kJ/mol), so make sure to multiply by the appropriate stoichiometric coefficients to account for the moles involved in the reaction.