Answer:
0.109 g.
Explanation:
Equation of the reaction:
Na3PO4 + 3HCl --> 3NaCl + H3PO4
Number of moles of HCl = molar concentration × volume
= 0.1 × 0.04
= 0.004 mol.
By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3
= 0.0013 mol
Mass of Na3PO4 = molar mass × number of moles
= 0.0013 × 164
= 0.219 g
Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance
= 0.219 × 50 g/100 g
= 0.109 g.
Answer:
Oh Okay
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The problem above can be solved using M1V1=M2V2 where M1 is the concentration of the concentrated, V1 is the volume of the concentrated solution, M2 is the concentration of the Dilute Solution, V2 is the Volume of the dilute solution. Hence,
(3.0 M)(V2)=(250 mL)(1.2M)
V2 (3.0)= 300
V2= 100 mL
Therefore, you need 100 mL of 3.0 M HCl to form a 250 mL of 1.2 M HCl.