Question

Assume that 5.60 \(LH_{2}\) at STP reacts with excess \(CuO\) according to the following equation: $$CuO(s)+H_{2}(g) \rightarrow Cu(s)+H_{2} O(g)$$ Make sure the equation is balanced before beginning your calculations. a. How many moles of \(H_{2}\) react? b. How many moles of \(Cu\) are produced? c. How many grams of \(Cu\) are produced?

Short answer

0.25 mol of \(H_{2}\) reacts, producing 0.25 mol of \(Cu\) or 15.89 grams of \(Cu\).

Step-by-step solution

Write down the balanced chemical equation

Ensure that the given equation is balanced. The provided equation, \[CuO(s) + H_{2}(g) \rightarrow Cu(s) + H_{2}O(g)\], is already balanced.

Calculate the moles of \(H_{2}\)

Use the ideal gas law (22.4 L/mol at STP) to convert the volume of \(H_{2}\) to moles. Given 5.60 L of \(H_{2}\), the moles of \(H_{2}\) can be calculated as follows: \[ \text{moles of } H_{2} = \frac{5.60 \text{ L}}{22.4 \text{ L/mol}} = 0.25 \text{ mol} \. \]

Determine the moles of \(Cu\) produced

From the balanced equation, the mole ratio of \(H_{2}\) to \(Cu\) is 1:1. Thus, the moles of \(Cu\) produced will be the same as the moles of \(H_{2}\) reacted: \[ \text{moles of } Cu = 0.25 \text{ mol} \. \]

Calculate the grams of \(Cu\) produced

Use the molar mass of copper (63.55 g/mol) to convert moles of \(Cu\) to grams: \[ \text{grams of } Cu = 0.25 \text{ mol} \times 63.55 \text{ g/mol} = 15.89 \text{ g} \. \]

Key concepts

balanced chemical equation

In stoichiometry, the first step is to ensure the chemical equation is balanced. This means the number of atoms for each element is the same on both the reactant and product sides of the equation. An unbalanced equation can lead to incorrect calculations. For the given reaction: \[ \text{CuO}(s) + \text{H}_2(g) \rightarrow \text{Cu}(s) + \text{H}_2\text{O}(g) \] check the number of atoms:

• 1 Cu atom on both sides
• 1 O atom on both sides
• 2 H atoms on both sides

The equation is already balanced. Now, we can move to the next steps confidently.

ideal gas law

The ideal gas law is crucial for converting gases between volume and moles, especially under standard temperature and pressure (STP). The law is given by: \[ PV = nRT \] Where:

• P is the pressure (at STP, P = 1 atm)
• V is the volume in liters
• n is the number of moles
• R is the gas constant (0.0821 L·atm/K·mol)
• T is the temperature in Kelvin (at STP, T = 273.15 K)

For simplicity, at STP, we use 22.4 L/mol as a quick conversion: \[ \text{moles} = \frac{V}{22.4 \text{ L/mol}} \] Using this relation, the volume of 5.60 L of \(\text{H}_2\) can be converted to: \[ \text{moles of } \text{H}_2 = \frac{5.60 \text{ L}}{22.4 \text{ L/mol}} = 0.25 \text{ mol} \] The calculated 0.25 moles of \(\text{H}_2\) can now be used for further calculations.

mole ratio

The mole ratio is derived from the coefficients of the balanced chemical equation. It tells us how reactants and products relate to each other in moles. For: \[ \text{CuO}(s) + \text{H}_2(g) \rightarrow \text{Cu}(s) + \text{H}_2\text{O}(g) \] The mole ratio between \(\text{H}_2\) and \(\text{Cu}\) is 1:1. This means one mole of \(\text{H}_2\) will produce one mole of \(\text{Cu}\). With our given problem: \[ \text{moles of } \text{H}_2: \text{Cu} = 1:1 \] Since we have 0.25 mol of \(\text{H}_2\), 0.25 mol of \(\text{Cu}\) will be produced.

molar mass

The molar mass is the mass of one mole of a substance. It allows us to convert between moles and grams. The molar mass of copper (\(\text{Cu}\)) is 63.55 g/mol. To find the grams of \(\text{Cu}\) produced, multiply the moles of \(\text{Cu}\) by its molar mass: \[ \text{grams of } \text{Cu} = 0.25 \text{ mol} \times 63.55 \text{ g/mol} \] This gives us: \[ \text{grams of } \text{Cu} = 15.89 \text{ g} \] In conclusion, 5.60 L of \(\text{H}_2\) at STP reacting with excess \(\text{CuO}\) produces 15.89 g of \(\text{Cu}\). This process involves ensuring the equation is balanced, using the ideal gas law, applying the mole ratio, and converting moles to grams using molar mass.