help please..?

Question

The standard enthalpy of combustion of methanol (CH3OH(l)) is -726 kJ·mol-1. How many kilojoules of heat are released for each 1.0 gram of methanol (CH3OH(l)) which is burned?

Answer 1

Find the quantity of methanol in moles:

n = m/Mr = 1/32 mol where Mr = 32 the relative molar mass of CH3OH.

Now when 1 mol of CH3OH is burned 726 kJ of heat are released
when 1/32 mol of CH3OH is burned q = 726/32 kJ of heat are released. So

q = 726/32 = 22.6875 kJ or 22.69 kJ (approx.)

Answer 2

1. The Mr of methanol is 32. Hence 32 g = 1 mole
2. Hence burning 32 g gives you 736 kJ
3. Therefore burning 1.0 g give you 736 kJ divided by 32....please work out the answer yourself

Answer 3

convert grams of methanol to moles of methanol (divide the grams of methanol (1) by the molecular mass of methanol (about 33, use the m0ost exact #'s you have for the mass of each atom) and multiply that by the heat of enthalpy (-726) (the moles will cancel, and your answer will be in kJ)

Answer 4

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Answer 5

kya yaar itna bhi nahi aata