Question

the volume of 0.25 m h3po4 required to neutralize 25ml of 0.03 Ca(OH)2 is​

Answer 1

Explanation:

Solution:- (B) 20 mL

2H3PO4+3Ca(OH)2⟶Ca3(PO4)2+6H2O

2 moles of H3PO4 reacts with 3 moles of Ca(OH)2

As we know that,

No. of moles =V(in L)×M

⇒V=Mno. of moles

whereas,

M = Molarity

V = Volume

Given that:-

Molarity of Ca(OH)2=0.03M

Volume of Ca(OH)2=25mL=2.5×10−2L

∴ No. of moles of Ca(OH)2=2.5×10−2×0.03=7.5×10−4 moles

No. of moles of H3PO4 required to react with 3 moles of CA(OH)2=2

No. of moles of H3

Answer 2

Answer:

Explanation:

equivalent = equivalent

0.5*v=1.5

v=3ml