Question

How many moles of \(Ba_3(PO_4)_2\) will be formed by the reaction of \(5\) moles of \(BaCl_2\) and \(3\) moles of \(Na_3(PO_4)\).

Answer 1

The chemical reaction between barium chloride (\(BaCl_2\)) and sodium phosphate (\(Na_3(PO_4)\)) is: \[ 3BaCl_2 + 2Na_3(PO_4) \rightarrow Ba_3(PO_4)_2 + 6NaCl \] Step 1: Determine the mole ratio from the balanced equation.
From the balanced equation, the mole ratio of \(BaCl_2\) to \(Ba_3(PO_4)_2\) is 3:1, and the mole ratio of \(Na_3(PO_4)\) to \(Ba_3(PO_4)_2\) is 2:1. 
Step 2: Compare the given amounts of reactants to the stoichiometric ratios.
We are given: - 5 moles of \(BaCl_2\) - 3 moles of \(Na_3(PO_4)\) 
Step 3: Find the limiting reactant.
For \(BaCl_2\), using the ratio of 3:1, we calculate the moles of \(Ba_3(PO_4)_2\) that can be formed: \[ \frac{5 \text{ moles of } BaCl_2}{3} = 1.67 \text{ moles of } Ba_3(PO_4)_2 \] For \(Na_3(PO_4)\), using the ratio of 2:1, we calculate the moles of \(Ba_3(PO_4)_2\) that can be formed: \[ \frac{3 \text{ moles of } Na_3(PO_4)}{2} = 1.5 \text{ moles of } Ba_3(PO_4)_2 \] The limiting reactant is \(Na_3(PO_4)\), as it can only produce 1.5 moles of \(Ba_3(PO_4)_2\). 
Final Answer:
Therefore, the number of moles of \(Ba_3(PO_4)_2\) formed is 1.5 moles.