m ) M How many milliliters of 0.750 M H2PO4 are required to react with 250. mL of 0.150 M Ba(OH)2 if the products are barium pho

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2 years ago

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sphate and water?

1 answer:

Inga [223] · Answer 1 · 2023-07-07T05:58:47+03:00

The volume of phosphoric acid is mathematically given as

X= 75 mL of H3PO4

<h3>What is the volume of H2PO4 required?</h3>

Question Parameters:

Milliliters of 0.750 M H2PO4 are required to react with 250. mL of 0.150 M Ba(OH)2

Generally, the equation for the banlanced Chemical Reaction is mathematically given as

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

Therefore

The volume of phosphoric acid= $\frac{3*2*250*150*3*1}{1*2*750}$

The volume of phosphoric acid=0.075 L of H3PO4

In conclusion, the volume of H2PO4 in mL is

The volume of phosphoric acid=75 mL of H3PO4