Respuesta :
Answer:
0.01125 moles of [tex]Ba_3(PO_4)_2(s)[/tex] can be produced from 115 mL of 0.218 M [tex]BaCl_2(aq)[/tex].
Explanation:
Moles of [tex]BaCl_2[/tex] = n
Volume of the solution = 115 mL = 0.115 L ( 1 mL=0.001 L)
Molarity of the [tex]BaCl_2[/tex] solution = 0.218 M
[tex]0.218 M=\frac{n}{0.115 L} [/tex]
[tex]n = 0.218\times 0.115 L=0.03379 mol[/tex]
[tex]3 BaCl_2(aq) + 2Na_3PO_4(aq)\rightarrow Ba_3(PO_4)_2(s) + 6NaCl(aq)[/tex]
According to reaction, 3 moles [tex]BaCl_2[/tex] gives 1 mole of [tex]Ba_3(PO_4)_2[/tex] .Then 0.03379 moles of [tex]BaCl_2[/tex] will give :
[tex]0.03379 mol\times \frac{1}{3}=0.01126 mol[/tex]
0.01125 moles of [tex]Ba_3(PO_4)_2(s)[/tex] can be produced from 115 mL of 0.218 M [tex]BaCl_2(aq)[/tex].
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
