Question

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)

Answer 1

Answer:

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

Answer 2

Answer:

0.01125 moles of $Ba_3(PO_4)_2(s)$ can be produced from 115 mL of 0.218 M $BaCl_2(aq)$.

Explanation:

Moles of $BaCl_2$ = n

Volume of the solution = 115 mL = 0.115 L ( 1 mL=0.001 L)

Molarity of the $BaCl_2$ solution = 0.218 M

$0.218 M=\frac{n}{0.115 L}$

$n = 0.218\times 0.115 L=0.03379 mol$

$3 BaCl_2(aq) + 2Na_3PO_4(aq)\rightarrow Ba_3(PO_4)_2(s) + 6NaCl(aq)$

According to reaction, 3 moles $BaCl_2$ gives 1 mole of $Ba_3(PO_4)_2$ .Then 0.03379 moles of $BaCl_2$ will give :

$0.03379 mol\times \frac{1}{3}=0.01126 mol$

0.01125 moles of $Ba_3(PO_4)_2(s)$ can be produced from 115 mL of 0.218 M $BaCl_2(aq)$.