methanol CH3OH is used as race car fuel write the balanced equation for the combustion of methanol. change in heat= -174kcal/mol methanol for the process How man kilocal are released u burning 33.0g of methanol
To write the balanced equation for the combustion of methanol (CH3OH), you need to know that methanol reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
The balanced equation is:
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O
Now, let's calculate the amount of heat released when 33.0 grams of methanol are burned.
1. Calculate the number of moles of methanol using the molar mass of methanol:
Molar mass of CH3OH = 12.01 g/mol (C) + 1.01 g/mol (H) + 16.00 g/mol (O) = 32.04 g/mol
Moles of CH3OH = 33.0 g / 32.04 g/mol = 1.03 mol
2. Use the molar ratio from the balanced equation to find the moles of heat released:
According to the balanced equation, 2 moles of CH3OH release -174 kcal/mol of heat.
Moles of heat released = 1.03 mol × (-174 kcal/mol) = -179.22 kcal
Therefore, burning 33.0 grams of methanol releases approximately -179.22 kilocalories of heat. Note that the negative sign indicates that heat is released (exothermic reaction).
To write the balanced equation for the combustion of methanol (CH3OH), we need to know that it reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced equation is as follows:
2 CH3OH + 3 O2 -> 2 CO2 + 4 H2O
From the balanced equation, we can find the molar ratio of methanol to energy released. Since the change in heat is given as -174 kcal/mol methanol, it means that 1 mol of methanol releases -174 kcal of heat energy.
To calculate how many kilocalories are released by burning 33.0g of methanol, we need to convert the grams of methanol to moles, and then multiply it by the heat released per mole.
Step 1: Calculate the number of moles of methanol.
The molar mass of methanol (CH3OH) is:
(1 x 12.01 g/mol) + (4 x 1.008 g/mol) + (1 x 16.00 g/mol) = 32.04 g/mol
Using the formula: moles = mass / molar mass,
moles = 33.0 g / 32.04 g/mol ≈ 1.03 mol
Step 2: Calculate the heat released by burning 33.0g of methanol.
Heat released = moles of methanol × heat released per mole
Heat released = 1.03 mol × -174 kcal/mol = -179.22 kcal
Therefore, by burning 33.0 g of methanol, approximately 179.22 kilocalories of energy are released.
2CH3OH + 3O2 => 2CO2 + 4H2O dH = -174 kcal/mol and that x 2 = -348 kcal/rxn and I'm taking you at your word that this is -174 kcal/mol and not -174 kcal/reaction for the 2 mols.
-348 kcal x 33.0/64 = ?