Question

click and drag on elements in order
during a reaction, 3.5 g of carbon monoxide gas (co)
reacts with an abundance of oxygen gas (o₂) to
produce carbon dioxide gas (co₂). all of the carbon
monoxide is used during the reaction. rank the order
of steps involved to determine the mass of carbon
dioxide (co₂) produced in this process.

instructions
0.1250 mol co₂ × (44.009g co₂)/(1 mol co₂) = 5.5 g co₂
2co(g) + o₂(g) → 2co₂(g)
0.1250 mol co × (2 mol co₂)/(2 mol co) = 0.1250 mol co₂
3.5 g co × (1 mol co)/(28.01g co) = 0.1250 mol co

Explanation:

Step1: Write the balanced equation

The balanced chemical equation for the reaction is \(2\text{CO}(g) + \text{O}_2(g)
ightarrow 2\text{CO}_2(g)\). This is the first step as we need the stoichiometric relationship from the balanced equation.

Step2: Convert mass of CO to moles

Using the molar mass of CO (28.01 g/mol), we convert the given mass (3.5 g) of CO to moles. The calculation is \(3.5\ \text{g CO} \times \frac{1\ \text{mol CO}}{28.01\ \text{g CO}} = 0.1250\ \text{mol CO}\).

Step3: Use stoichiometry to find moles of \(CO_2\)

From the balanced equation, the mole ratio of \(CO\) to \(CO_2\) is 2:2 (or 1:1). So we calculate moles of \(CO_2\) as \(0.1250\ \text{mol CO} \times \frac{2\ \text{mol CO}_2}{2\ \text{mol CO}} = 0.1250\ \text{mol CO}_2\).

Step4: Convert moles of \(CO_2\) to mass

Using the molar mass of \(CO_2\) (44.009 g/mol), we convert moles of \(CO_2\) to mass. The calculation is \(0.1250\ \text{mol CO}_2 \times \frac{44.009\ \text{g CO}_2}{1\ \text{mol CO}_2} = 5.5\ \text{g CO}_2\).

Answer:

1. \(2\text{CO}(g) + \text{O}_2(g)

ightarrow 2\text{CO}_2(g)\)

1. \(3.5\ \text{g CO} \times \frac{1\ \text{mol CO}}{28.01\ \text{g CO}} = 0.1250\ \text{mol CO}\)
2. \(0.1250\ \text{mol CO} \times \frac{2\ \text{mol CO}_2}{2\ \text{mol CO}} = 0.1250\ \text{mol CO}_2\)
3. \(0.1250\ \text{mol CO}_2 \times \frac{44.009\ \text{g CO}_2}{1\ \text{mol CO}_2} = 5.5\ \text{g CO}_2\)