<u>Answer:</u> The mass of methanol that must be burned is 24.34 grams
<u>Explanation:</u>
We are given:
Amount of heat produced = 581 kJ
For the given chemical equation:
![CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ](https://tex.z-dn.net/?f=CH_3OH%28g%29%2B%5Cfrac%7B3%7D%7B2%7DO_2%28g%29%5Crightarrow%20CO_2%28g%29%2B2H_2O%28l%29%3B%5CDelta%20H%3D-764kJ)
By Stoichiometry of the reaction:
When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole
So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = ![\frac{1}{764}\times 581=0.7605mol](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B764%7D%5Ctimes%20581%3D0.7605mol)
To calculate mass for given number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Moles of methanol = 0.7605 moles
Molar mass of methanol = 32 g/mol
Putting values in above equation, we get:
![0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g](https://tex.z-dn.net/?f=0.7605mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20methanol%7D%7D%7B32g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20methanol%7D%3D%280.7605mol%5Ctimes%2032g%2Fmol%29%3D24.34g)
Hence, the mass of methanol that must be burned is 24.34 grams