Questions: According to the following reaction, how many grams of sulfur trioxide will be formed upon the complete reaction of 21.4 grams of oxygen gas with excess sulfur dioxide? sulfur dioxide (g)+ oxygen (g) → sulfur trioxide (g) Mass = grams sulfur trioxide

According to the following reaction, how many grams of sulfur trioxide will be formed upon the complete reaction of 21.4 grams of oxygen gas with excess sulfur dioxide?
sulfur dioxide (g)+ oxygen (g) → sulfur trioxide (g)
Mass = grams sulfur trioxide

Transcript text: According to the following reaction, how many grams of sulfur trioxide will be formed upon the complete reaction of 21.4 grams of oxygen gas with excess sulfur dioxide? sulfur dioxide $(g)+$ oxygen $(g) \rightarrow$ sulfur trioxide $(g)$ Mass = $\square$ grams sulfur trioxide

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The given reaction is between sulfur dioxide ($\text{SO}_2$) and oxygen ($\text{O}_2$) to form sulfur trioxide ($\text{SO}_3$). The balanced chemical equation is:

\[ 2 \, \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \, \text{SO}_3(g) \]

Step 2: Calculate Moles of Oxygen Gas

First, we need to calculate the number of moles of oxygen gas ($\text{O}_2$) using its molar mass. The molar mass of $\text{O}_2$ is approximately 32.00 g/mol.

\[ \text{Moles of } \text{O}_2 = \frac{21.4 \, \text{g}}{32.00 \, \text{g/mol}} = 0.6688 \, \text{mol} \]

Step 3: Use Stoichiometry to Find Moles of Sulfur Trioxide

From the balanced equation, 1 mole of $\text{O}_2$ produces 2 moles of $\text{SO}_3$. Therefore, 0.6688 moles of $\text{O}_2$ will produce:

\[ 0.6688 \, \text{mol} \times 2 = 1.3376 \, \text{mol of } \text{SO}_3 \]

Step 4: Calculate Mass of Sulfur Trioxide

The molar mass of sulfur trioxide ($\text{SO}_3$) is approximately 80.06 g/mol. Using this, we can calculate the mass of $\text{SO}_3$ formed:

\[ \text{Mass of } \text{SO}_3 = 1.3376 \, \text{mol} \times 80.06 \, \text{g/mol} = 107.08 \, \text{g} \]