Liquid methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be used as an alternative fuel by pickup trucks and SUVs. An industrial method for preparing it involves the catalytic hydrogenation of carbon monoxide: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \stackrel{\text { catalyss }}{\longrightarrow} \mathrm{CH}_{3} \mathrm{OH}(l)$$How much heat (in \(\mathrm{kJ}\) ) is released when \(15.0 \mathrm{~L}\) of \(\mathrm{CO}\) at \(85^{\circ} \mathrm{C}\) and \(112 \mathrm{kPa}\) reacts with \(18.5 \mathrm{~L}\) of \(\mathrm{H}_{2}\) at \(75^{\circ} \mathrm{C}\) and 744 torr?

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and the number of moles of a gas. The formula is expressed as: \[ PV = nRT \] where:

• P is the pressure of the gas in kilopascals (kPa) or other units.
• V is the volume of the gas in liters (L).
• n is the number of moles of the gas.
• R is the ideal gas constant, which is 8.314 kPa·L·K^{-1}·mol^{-1}.
• T is the temperature in Kelvin (K).

Temperature must always be in Kelvin. To convert from Celsius to Kelvin, add 273.15. In the given exercise, you used this law to convert the volumes of CO and H2 into moles. For CO:
Volume (V) = 15.0 L
Pressure (P) = 112 kPa
Temperature (T) = 358.15 K
You then plugged these values into the equation,
\[ n_{CO} = \frac{PV}{RT} = \frac{112 \text{kPa} \times 15.0 \text{L}}{8.314 \text{kPa·L·K}^{-1}\text{·mol}^{-1} \times 358.15 \text{K}} = 0.563 \text{mol} \] Similarly, the values were converted for H2:
Volume (V) = 18.5 L
Pressure (P) = 99.19 kPa
Temperature (T) = 348.15 K.
\[ n_{H2} = \frac{PV}{RT} = \frac{99.19 \text{kPa} \times 18.5 \text{L}}{8.314 \text{kPa·L·K}^{-1}\text{·mol}^{-1} \times 348.15 \text{K}} = 0.635 \text{mol} \] This law allows us to relate the given conditions to find the amount of gas in moles.

The concept of the limiting reactant is crucial in chemistry. It determines the maximum amount of product that can be formed in a reaction. In a chemical reaction, the limiting reactant is the reactant that is completely consumed first, thus limiting the extent of the reaction. To identify the limiting reactant, follow these steps:

• Write the balanced chemical equation.
• Calculate the moles of each reactant.
• Use the stoichiometry (mole ratios) from the balanced equation to determine which reactant will run out first.

In the exercise, the balanced chemical equation is:
\[ \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(l) \]From Step 1:
\[ n_{CO} = 0.563 \text{mol} \]
\[ n_{H2} = 0.635 \text{mol} \] According to the balanced equation, 1 mole of CO reacts with 2 moles of H2. Therefore, to react with 0.563 mol of CO, we need: \[ 2 \times 0.563 \text{mol} = 1.126 \text{mol of } \mathrm{H}_{2} \] Given that only 0.635 mol of H2 is available, we do not have enough H2 to completely react with all of the CO. This makes CO the limiting reactant as it gets completely consumed first. The remaining H2 will be in excess.

Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction. It uses the balanced chemical equation to find the relationships between the moles of reactants and products. The coefficients in the balanced equation indicate the mole ratios of each substance involved. Here's a breakdown of how it's applied:

• Write the balanced chemical equation.
• Determine the mole ratios.
• Calculate the moles of desired reactants or products using the relationships from the balanced equation.

For our reaction:
\[ \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(l) \] The mole ratio between CO and H2 is 1:2. This means 1 mole of CO reacts with 2 moles of H2. With CO as limiting, only 0.563 mol of CO can fully react. Stoichiometry helps us predict that this will produce:
\[ n_{\mathrm{CH}_{3}\mathrm{OH}} = 0.563 \text{mol} \]Since the reaction is exothermic, it releases energy. The enthalpy change (ΔH) for the formation of methanol is:
\[ \Delta H = -90.7 \text{ kJ/mol} \] The heat released when 0.563 mol of CO reacts to form methanol is: \[ \text{Heat released} = \Delta H \times \text{moles of CO} = -90.7 \text{kJ/mol} \times 0.563 \text{mol} \approx -51.0 \text{kJ} \] This illustrates how stoichiometry, using the balanced equation and mole ratios, allows us to understand the quantities involved in the reaction and predict the energy change.