According to the reaction below, how many moles of Ba3(PO4)2(s) can be produced from 115 mL of 0.218 M BaCl2(aq)? Assume that there is excess Na3PO4(aq).

3 BaCl2(aq) + 2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaCl(aq)

We can produce 0.00836 moles of Ba3(PO4)2

Explanation:

Step 1: Data given

Volume BaCl2 = 115 mL = 0.115 L

Molarity BaCl2 = 0.218 M

Step 2: The balanced equation

3 BaCl2(aq) + 2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaCl(aq)

Step 3: Calculate moles BaCl2

Moles BaCl2 = molarity * volume

Moles BaCl2 = 0.218 M * 0.115 L

Moles BaCl2 = 0.02507 moles

Step 4: Calculate moles Ba3(PO4)2

For 3 moles BaCl2 we need 2 moles Na3PO4 to produce 1 mol Ba3(PO4)2 and 6 moles NaCl

For 0.02507 moles BaCl2 we'll have 0.02507/3 = 0.00836 moles

We can produce 0.00836 moles of Ba3(PO4)2

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