Assume that the reaction $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ goes to completion. When 10. mol of \(\mathrm{CO}\) and \(8.0 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) react in a closed 10.-L vessel, (a) how many moles of \(\mathrm{CO}, \mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) are present at the end of the reaction? (b) what will be the total pressure in the flask at \(0^{\circ} \mathrm{C}\) ?

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, limiting the amount of product that can be formed. To identify the limiting reactant, we compare the mole ratio of each reactant available to the ratio required by the balanced chemical equation.

In the given problem, the balanced equation is:

$$2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2$$

There are 10 moles of CO and 8 moles of O2. We use the coefficients from the balanced equation to determine which reactant will run out first. For CO, the calculation is:

\frac{10 \text{ moles CO}}{2 \text{ moles CO}} = 5\ If we need 2 moles of CO for every mole of O2, let's see how much O2 is actually available:

$\frac{8 \text{ moles O2}}{1 \text{ mole O2}} = 8\ Since 5 < 8, CO will be the limiting reactant. Once all 10 moles of CO are used, the reaction stops, regardless of how much O2 is left.

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas. It's an essential tool in chemistry for finding one of these variables when the others are known:

$$PV = nRT$$

Here, P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

In this problem, we aim to find the final pressure in the flask. We know:

• Total volume (V) = 10 L
• Temperature (T) = 273 K (0°C)
• Total moles (n) = moles CO2 + moles O2 = 10 + 3 = 13 moles
• Gas constant (R) = 0.0821 L·atm/(K·mol)

We substitute these values into the Ideal Gas Law:

\frac{nRT}{V}\ = \frac{13 \times 0.0821 \times 273}{10} = 29.1 \text{ atm} \ So, the total pressure in the flask at 0°C is 29.1 atm.

Stoichiometry is the calculation of reactants and products in chemical reactions. It's based on the conservation of mass and the ratios provided by the balanced chemical equation.

For the reaction:

$$2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2$$

The coefficients tell us that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2. These coefficients can be used to determine how much product we get from a known amount of reactants or how much of one reactant is needed to fully react with another.

In the problem, we start with 10 moles of CO and 8 moles of O2. By identifying CO as the limiting reactant, we can use stoichiometry to calculate the products and unused reactants:

- For every 2 moles of CO used, 1 mole of O2 is required.

- 10 moles of CO will thus use up \frac{10}{2} = 5\text{ moles O2}\

- 10 moles of CO produce 10 moles of CO2.

Remaining O2: \frac{8 - 5}{1} = 3 \text{ moles}\ Thus, products and remaining reactants are:

• 0 moles of CO
• 3 moles of O2
• 10 moles of CO2

Mole ratio is crucial in solving stoichiometric problems. It refers to the ratio of moles of one substance to the moles of another substance in a balanced equation.

In our equation:

$$2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2$$

The mole ratios are:

• 2 moles of CO to 1 mole of O2
• 2 moles of CO to 2 moles of CO2
• 1 mole of O2 to 2 moles of CO2

These ratios help in calculating the required or produced quantities in a reaction.

For example, if we need to know how much O2 is required to react with 10 moles of CO, we use the mole ratio between CO and O2:

10 moles of CO * 1 mole of O2 / 2 moles of CO = 5 moles of O2

Understanding mole ratios helps us quickly scale reactions up or down depending on how much of each compound we start with or need to produce.