What mass of solid magnesium hydroxide is needed to neutralize .385 L of 2.5 M nitric acid?
2HNO3+Mg(OH)2>>Mg(NO3)2 + 2 H2O
so you need 1/2 mole magnesium hydroxide for each mole of nitric acid.
MoleNitricAcid=.385*2.5*(molmassNitricacid)
= .385*2.5*63
MassMg(OH)2=above/2 * molmassMgOH2
= above/2 * 58.3 grams
Mg(OH)2 + 2HNO3 => Mg(NO3)2 + 2HOH
moles = Molarity x Volume(L)
moles HNO3 = (2.5M)(0.385L) =(2.5mol/L)(0.385L)= 0.963mole HNO3
From balanced equation, half as many moles of Mg(OH)2 are needed as the amount of HNO3 given. That is, moles of Mg(OH)2 needed is 1/2(0.963 mole) = 0.481 mole Mg(OH)2 used to neutralize 0.963mole of HNO3. Converting moles to grams by multiplying by formula wt of Mg(OH)2 (= 58.33 g/mol) => (0.481mol)(58.33 g/mol) = 28.1 gms of Mg(OH)2 needed to neutralize .385L of 2.5M HNO3.
To find the mass of solid magnesium hydroxide needed to neutralize the given volume and concentration of nitric acid, we need to use stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation between magnesium hydroxide (Mg(OH)2) and nitric acid (HNO3) is:
2 HNO3 + Mg(OH)2 -> Mg(NO3)2 + 2H2O
From the balanced equation, we can see that one mole of magnesium hydroxide reacts with two moles of nitric acid. This indicates a 1:2 ratio between magnesium hydroxide and nitric acid.
First, let's calculate the number of moles of nitric acid in the given volume of 0.385 L and concentration of 2.5 M:
moles of HNO3 = concentration × volume
= 2.5 M × 0.385 L
= 0.9625 moles
Since the ratio between magnesium hydroxide and nitric acid is 1:2, we need half the moles of magnesium hydroxide to neutralize the nitric acid:
moles of Mg(OH)2 = 0.9625 moles ÷ 2
= 0.48125 moles
Finally, to find the mass of magnesium hydroxide, we need to use its molar mass. The molar mass of magnesium hydroxide, Mg(OH)2, is 58.33 g/mol.
Mass of Mg(OH)2 = moles × molar mass
= 0.48125 moles × 58.33 g/mol
= 28.06 grams (rounded to two decimal places)
Therefore, approximately 28.06 grams of solid magnesium hydroxide is needed to neutralize 0.385 L of 2.5 M nitric acid.