The molarity of the solution of H₃PO₄ needed to neutralize the KOH solution is 0.35 M
<h3>Balanced equation </h3>
H₃PO₄ + 3KOH —> K₃PO₄ + 3H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 1
- The mole ratio of the base, KOH (nB) = 3
<h3>How to determine the molarity of H₃PO₄ </h3>
- Volume of acid, H₃PO₄ (Va) = 10.2 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.2 M
- Volume of base, Ca(OH)₂ (Vb) = 53.5 mL
- Molarity of acid, H₃PO₄ (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 10.2) / (0.2 × 53.5) = 1 / 3
(Ma × 10.2) / 10.7 = 1 / 3
Cross multiply
Ma × 10.2 × 3 = 10.7
Ma × 30.6 = 10.7
Divide both side by 30.6
Ma = 10.7 / 30.6
Ma = 0.35 M
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Each element is diffrent than another element
The balanced chemical reaction would be:
FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)
We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:
Amount of HCl used: 0.240 mol FeS x 2 mol HCl / 1 mol FeS = 0.48 mol HCl
0.646 - 0.48 = 0.166 mol HCl left
