3.8mL of 0.42 phosphoric acid is required.
Reaction
2H3PO4 + 3CaCL3 → Ca3(PO)4 + 6HCl
moles CaCl2 =0.16 mol/L x0.010 L = 0.0016 mol
moles of H3Po4
= 0.0016mol of CaCl2 x 2 mole of H3PO4/3mole of CaCl2
= 0.00106 mol
V of H3PO4 = 0.0016/0.42 = 0.0038L = 3.8mL
V of H3PO4=3.8mL
To know more about calculation in milliliters refer to:-
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