Question

How many grams of CO are produced when 33.0 g of C reacts? Fe2O3(s)+3C(s)→2Fe(s)+3CO(g)

Answer 1

To calculate the mass of CO produced from 33.0 g of C, use stoichiometry and molar masses to convert from grams of C to moles and then to grams of CO, resulting in approximately 77.0 g of CO.

Step-by-step explanation:

To calculate the mass of CO produced when 33.0 g of C reacts according to the balanced chemical equation Fe2O3(s) + 3C(s) → 2Fe(s) + 3CO(g), you must first determine the molar mass of carbon (C) and carbon monoxide (CO), and use stoichiometry to convert grams of C to moles, and then use the molar ratio from the balanced equation to find the moles of CO produced.

First, calculate the moles of C by dividing the mass of carbon by its molar mass (approximately 12.01 g/mol). Then, use the molar ratio from the balanced equation (3 mol C to 3 mol CO) to find the moles of CO produced. Finally, multiply the moles of CO by its molar mass (approximately 28.01 g/mol) to find the mass of CO produced.

Step-by-step calculation:

1. Moles of C = 33.0 g / 12.01 g/mol = 2.748 moles C

2. The molar ratio is 3:3, so moles of CO = moles of C = 2.748 moles CO

3. Mass of CO = moles of CO x 28.01 g/mol = 76.96 g CO

The mass of CO produced is therefore approximately 76.96 grams.

The calculation results are typically reported to three significant figures, so the final answer would be 77.0 g of CO produced.

Answer 2

77.0 grams of CO are produced when 33.0 g of C reacts.

Explanation

Given:

Mass of C = 33.0 g

Equation: Fe₂O₃(s) + 3C(s) → 2Fe(s) + 3CO(g)

What to find;

The grams of CO that are produced when 33.0 g of C reacts.

Step-by-step solution:

Step 1: Convert 33.0 g of C to moles.

Using the formula for moles and the molar mass of C = 12.0 g/mol, the moles of C is

`\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ Moles\text{ }of\text{ }C=\frac{33.0\text{ }g}{12.0\text{ }g\text{/}mol}=2.75\text{ }mol \end{gathered}`

Step 2: Determine the moles of CO produced.

Using the mole ratio of C and CO from the balance equation and the moles of C in step 1; the moles of CO produced is calculated as follows:

`\begin{gathered} 3\text{ }mol\text{ }C=3\text{ }mol\text{ }CO \\ \\ 2.75\text{ }mol\text{ }C=x \\ \\ x=\frac{2.75\text{ }mol\text{ }C}{3\text{ }mol\text{ }C}*3\text{ }mol\text{ }CO \\ \\ x=2.75\text{ }mol\text{ }CO \end{gathered}`

Step 3: Convert the moles of CO produced to mass in grams.

The moles of CO produced in step above can be converted to mass in grams as shown below

`\begin{gathered} Mass=Moles* Molar\text{ }mass \\ \\ Molar\text{ }mass\text{ }of\text{ }CO=12.0+16.0=28.0\text{ }g\text{ /}mol \\ \\ Mass\text{ }of\text{ }CO\text{ }produced=2.75\text{ }mol*28.0\text{ }g\text{/}mol \\ \\ Mass\text{ }of\text{ }CO\text{ }produced=77.0\text{ }g \end{gathered}`

Hence, 77.0 grams of CO are produced when 33.0 g of C reacts.