Question

question 1
11.08 g of liquid methanol (ch₃oh) is burned in excess oxygen gas according to the following reaction:
2 ch₃oh(l) + 3 o₂(g) → 2 co₂(g) + 4 h₂o(l)
the reaction releases 251 kj of heat. what is the δh for the reaction as written?
kj/mol
question 2
calculate the amount of heat produced when 4.414 l of liquid acetonitrile (c₂h₃n) is burned in excess oxygen gas at 25°c according to the following thermochemical reaction:
4 c₂h₃n(l) + 11 o₂(g) → 8 co₂(g) + 6 h₂o(g) + 2 n₂(g) δh = -4989 kj
the density of acetonitrile at 25°c is 0.7857 g/ml.
kj

Explanation:

Step1: Calculate the molar mass of \(CH_3OH\)

The molar mass of \(C = 12.01\space g/mol\), \(H= 1.008\space g/mol\), \(O = 16.00\space g/mol\).
For \(CH_3OH\), \(M=(12.01)+(4\times1.008)+16.00 = 32.042\space g/mol\)

Step2: Calculate the number of moles of \(CH_3OH\)

Using the formula \(n=\frac{m}{M}\), where \(m = 11.08\space g\) and \(M = 32.042\space g/mol\)
\(n=\frac{11.08}{32.042}\approx0.346\space mol\)

Step3: Relate the heat released to \(\Delta H\)

From the balanced equation, \(2\space mol\) of \(CH_3OH\) reacts.
If \(0.346\space mol\) of \(CH_3OH\) releases \(251\space kJ\) of heat, then for \(2\space mol\) of \(CH_3OH\)
\(\Delta H=\frac{251}{0.346}\times2\approx - 1450\space kJ/mol\) (negative because heat is released)

Step1: Convert volume of acetonitrile to mass

Given \(V = 4.414\space L=4414\space mL\) and \(
ho=0.7857\space g/mL\)
Using \(m=
ho V\), \(m = 0.7857\times4414 = 3469.9\space g\)

Step2: Calculate the molar mass of \(C_2H_3N\)

The molar mass of \(C = 12.01\space g/mol\), \(H = 1.008\space g/mol\), \(N=14.01\space g/mol\)
For \(C_2H_3N\), \(M=(2\times12.01)+(3\times1.008)+14.01=53.064\space g/mol\)

Step3: Calculate the number of moles of \(C_2H_3N\)

Using \(n=\frac{m}{M}\), \(n=\frac{3469.9}{53.064}\approx65.4\space mol\)

Step4: Calculate the heat produced

From the balanced equation, \(4\space mol\) of \(C_2H_3N\) gives \(\Delta H=- 4989\space kJ\)
For \(n = 65.4\space mol\) of \(C_2H_3N\)
\(q=\frac{65.4}{4}\times(-4989)\approx - 81600\space kJ\)

Answer:

\(-1450\)

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