Question

we wish to determine how many grams of al(no₃)₃ can form when 200.0 ml of 0.500 m al₂(so₄)₃ reacts with excess ba(no₃)₂. 3ba(no₃)₂(aq) + al₂(so₄)₃(aq) → 3baso₄(s) + 2al(no₃)₃(aq) how many moles of al₂(so₄)₃ are present in 200.0 ml of 0.500 m al₂(so₄)₃?

Explanation:

Step1: Recall the formula for moles from molarity

The formula for calculating moles (\(n\)) from molarity (\(M\)) and volume (\(V\)) is \(n = M\times V\), where the volume should be in liters.

Step2: Convert volume to liters

Given volume \(V = 200.0\space mL\). Since \(1\space L = 1000\space mL\), we convert \(200.0\space mL\) to liters: \(V=\frac{200.0}{1000}=0.2000\space L\).

Step3: Calculate moles of \(Al_2(SO_4)_3\)

Given molarity \(M = 0.500\space M\) and volume \(V = 0.2000\space L\). Using the formula \(n = M\times V\), we substitute the values: \(n=0.500\space mol/L\times0.2000\space L = 0.100\space mol\).

Answer:

The number of moles of \(Al_2(SO_4)_3\) is \(0.100\space mol\).