Question

when methanol, ch₃oh, is burned in the presence of oxygen gas, o₂, a large amount of heat energy is released. for this reason, it is often used as a fuel in high performance racing cars. the combustion of methanol has the balanced, thermochemical equation
ch₃oh(g) + ³⁄₂o₂(g) → co₂(g) + 2h₂o(l) δh = -764 kj
how much methanol, in grams, must be burned to produce 857 kj of heat?
mass: g

Explanation:

Step1: Calculate the moles of methanol

From the thermochemical equation, when \(1\) mole of \(CH_3OH\) is burned, \(\Delta H=- 764\space kJ\). Let \(n\) be the moles of \(CH_3OH\). We know that \(\frac{n}{1}=\frac{857}{764}\) (since the heat released is proportional to the moles of methanol burned). So \(n=\frac{857}{764}\space mol\)

Step2: Calculate the molar mass of methanol

The molar mass of \(CH_3OH\): \(M=(12 + 1\times4+16)\space g/mol=32\space g/mol\)

Step3: Calculate the mass of methanol

Using the formula \(m = n\times M\), substitute \(n=\frac{857}{764}\space mol\) and \(M = 32\space g/mol\). Then \(m=\frac{857}{764}\times32\space g\)

$$m=\frac{857\times32}{764}\space g=\frac{27424}{764}\space g = 35.9\space g$$

Answer:

\(35.9\space g\)