Questions: Which substance is in excess when 1.50 mol of aluminum bromide is reacted with 2.50 mol of barium hydroxide in the following equation? A) BaBr2 B) Al(OH)3 C) Ba(OH)2 D) AlBr3

Which substance is in excess when 1.50 mol of aluminum bromide is reacted with 2.50 mol of barium hydroxide in the following equation?
A) BaBr2
B) Al(OH)3
C) Ba(OH)2
D) AlBr3

Transcript text: Which substance is in excess when 1.50 mol of aluminum bromide is reacted with 2.50 mol of barium hydroxide in the following equation? A) BaBr 2 B) $\mathrm{Al}(\mathrm{OH}) 3$ C) $\mathrm{Ba}(\mathrm{OH}) 2$ D) AlBr 3

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the reaction between aluminum bromide ($\text{AlBr}_3$) and barium hydroxide ($\text{Ba(OH)}_2$) is:

\[ 2 \text{AlBr}_3 + 3 \text{Ba(OH)}_2 \rightarrow 3 \text{BaBr}_2 + 2 \text{Al(OH)}_3 \]

Step 2: Determine the Mole Ratio

From the balanced equation, the mole ratio of $\text{AlBr}_3$ to $\text{Ba(OH)}_2$ is 2:3.

Step 3: Calculate the Required Moles

To find out which reactant is in excess, calculate the moles of $\text{Ba(OH)}_2$ required to completely react with 1.50 mol of $\text{AlBr}_3$:

\[ \text{Moles of } \text{Ba(OH)}_2 = \frac{3}{2} \times 1.50 = 2.25 \text{ mol} \]

Step 4: Compare with Available Moles

We have 2.50 mol of $\text{Ba(OH)}_2$ available, which is more than the 2.25 mol required. Therefore, $\text{Ba(OH)}_2$ is in excess.

Final Answer

$\boxed{\text{C) Ba(OH)}_2}$

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Questions: Which substance is in excess when 1.50 mol of aluminum bromide is reacted with 2.50 mol of barium hydroxide in the following equation? A) BaBr2 B) Al(OH)3 C) Ba(OH)2 D) AlBr3

Solution

Solution Steps

Final Answer

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