Respuesta :
Answer:
Explanation:
2HNO3 + Ca(OH)2 ---> Ca(NO3)2 + 2H2O
moles HNO3---> (0.2545 mol/L) (0.04587 L) = 0.011674 mol
use the 2 : 1 molar ratio of HNO3 to Ca(OH)2:
2 is to 1 as 0.011674 mol is to x
x = 0.005837 mol of Ca(OH)2
molarity of Ca(OH)2 ---> 0.005837 mol / 0.02162 L = 0.2700 M (to four sig figs)
Answer:
The volume of 0.100 M H₂SO₄ needed to neutralize the resulting mixture containing an excess of 10 ml of 0.100 M Ca(OH)₂ is 10 ml of 0.100 M H₂SO₄ .
Explanation:
Ca(OH)₂ + HNO₃ = Ca(NO₃)₂ + H₂O
From the above reaction;
1 mole of Ca(OH)₂ is required to neutralize 1 mole of HNO₃ also
0.1 mole of Ca(OH)₂ will be required to neutralize 0.1 mole of HNO₃
Since there is 60 ml of 0.100 M Ca(OH)₂ and 50.0 mL of 0.100 M HNO₃, then on completion of neutralization, there will be 10 ml of 0.100 M Ca(OH)₂ left to be consumed
When H₂SO₄ is added to the solution, the reaction will be as follows;
Ca(OH)₂ + H₂SO₄ → CaSO₄ + H₂O
Therefore we have;
1 mole of H₂SO₄ is required to completely react with 1 mole of Ca(OH)₂
Hence,
0.1 mole of H₂SO₄ will completely react with 0.1 mole of Ca(OH)₂
Therefore the volume of 0.100 M H₂SO₄ required to neutralize the resulting mixture containing an excess of 10 ml of 0.100 M Ca(OH)₂ is 10 ml
