If anyone could please help me with this i'd really appreciate it!

Consider the following acid/base reaction: 2H3PO4 (aq) + 3 Ba(OH)2 (aq) ---> 6H2O (l) + Ba3(PO4)2 (s)

The titration of an unknown concentration H3PO4 solution requires 0.186 L of 0.214 M Ba(OH)2 solution. How any moles of H3PO4 are determined from this procedure?

Question

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joseaaronlara joseaaronlara · Answer 1 · 2019-10-21T06:02:30+02:00

Answer:

The answer to your question is: 0.0265 moles of H3PO4

Explanation:

Data

[H3PO4] = ?

Volume Ba(OH)2= 0.186 l

[Ba(OH)2] = 0.214 M

Equation

2H3PO4 (aq) + 3 Ba(OH)2 (aq) ---> 6H2O (l) + Ba3(PO4)2 (s)

Process

Molarity = moles / volume

moles = Molarity x volume

moles of Ba(OH)2 = 0.214 x 0.186

= 0.0398

Rule of three

2 moles of H3PO4 -------------------- 3 moles of Ba(OH)2

x -------------------- 0.0398 moles of Ba(OH)2

x = (0.0398 x 2) / 3

x = 0.0265 moles of H3PO4

Аноним Аноним · Answer 2 · 2019-10-21T13:31:08+02:00

Answer:

0.027 moles to the nearest thousandth.

Explanation:

2H3PO4 (aq) + 3 Ba(OH)2 (aq) ----> 6H2O (l) + Ba3(PO4)2 (s)

From the equation 3 moles of Ba(OH)2 reacts with 2 moles of H3PO4.

So 1 mole of Ba(OH)2 reacts with 2/3 mole of H3PO4.

The number of moles of Ba(OH)2 in the solution is 0.214 * 0.168 = 0.039804 moles.

So the number of moles of H3PO4 is 2/3 * 0.039804

= 0.027 moles to the nearest thousandth.