What quantity of energy is released when 506 g of liquid water freezes?
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1a. The balanced equation for the reaction is:
<h3>3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O </h3>1b. The number of mole of Ca(OH)₂ is 0.0247 mole
1c. The number of mole of H₃PO₄ is 0.0165 mole.
1d. The concentration of H₃PO₄ is 0.432 mol/L
2. The new concentration of the H₃PO₄ solution is 0.0432 mol/L
<h3>1a. The balanced equation for the reaction</h3><u>3</u>Ca(OH)₂ + <u>2</u>H₃PO₄ —> Ca₃(PO₄)₂ + <u>6</u>H₂O
<h3>1b. Determination of the mole of Ca(OH)₂</h3>Volume of Ca(OH)₂ = 71 mL = 71 / 1000 = 0.071 L
Concentration of Ca(OH)₂ = 0.348 mol/L
<h3>Mole of Ca(OH)₂ =? </h3>Mole = Concentration × Volume
Mole = 0.348 × 0.071
<h3>Mole of Ca(OH)₂ = 0.0247 mole </h3><h3>1c. Determination of the mole of H₃PO₄. </h3>3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
3 moles of Ca(OH)₂ reacted with 2 moles of H₃PO₄.
Therefore,
0.0247 moles of Ca(OH)₂ will react with = = 0.0165 mole of H₃PO₄.
Thus, the number of mole of H₃PO₄ is 0.0165 mole
<h3>1d. Determination of the concentration of H₃PO₄</h3>Volume of H₃PO₄ = 38.20 mL = 38.20/ 1000 = 0.0382 L
Mole of H₃PO₄ = 0.0165 mole
<h3>Concentration of H₃PO₄ =?</h3>Initial Volume (V₁) = 10 mL
Initial concentration (C₁) = 0.432 mol/L
New volume (V₂) = 100 mL
<h3>New concentration (C₂) =?</h3>The new concentration of the H₃PO₄ solution can be obtained as follow:
<h3>C₁V₁ = C₂V₂</h3>0.432 × 10 = C₂ × 100
4.32 = C₂ × 100
Divide both side by 100
C₂ =
Therefore, the new concentration of the H₃PO₄ solution is 0.0432 mol/L
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