Answer:
THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.
Explanation:
Using
Ca VA / Cb Vb = Na / Nb
Ca = 0.0500 M
Va = 35 mL
Cb = 0.0200 M
Vb = unknown
Na = 2
Nb = 1
Equation for the reaction:
Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O
So therefore, we make Vb the subject of the equation and solve for it
Vb = Ca Va Nb / Cb Na
Vb = 0.0500 * 35 * 1 / 0.0200 * 2
Vb = 1.75 / 0.04
Vb = 43.75 mL
The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL
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Mole ratio:
MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl
2 moles KOH ---------------- 1 mole Mg(OH)₂
4 moles KOH ------------------- moles Mg(OH)₂
moles Mg(OH)₂ = 4 x 1 / 2
= 2 moles of Mg(OH)₂
molar mass Mg(OH)₂ = 58g/mol
mass of Mg(OH)₂ = n x mm
mass of Mg(OH)₂ = 58 x 2
= 116 g of <span> Mg(OH)₂</span>
hope this helps!