In an experiment to determine the percentage of magnesium hydroxide in an anti-acid,a solution containing 0.50g of the anti-acid was neutralized by 23.0of 0.010M hydrochloric acid(R.F.M of magnesium hydroxide=58)
a)Mass of magnesium hydroxide in the anti-acid
b) Percentage of magnesium hydroxide in the anti-acid
c)When 8.53g of sodium nitrate in an open tube the mass of oxygen gas produced was 0.83g.Given the equation as 2NaNo3~2NaNo2+O2
Calculate the percentage of sodium nitrate that was converted to sodium nitrite
a) To determine the mass of magnesium hydroxide in the anti-acid, use the balanced chemical equation between magnesium hydroxide and hydrochloric acid:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
From the equation, we can see that 1 mole of magnesium hydroxide reacts with 2 moles of hydrochloric acid.
First, calculate the number of moles of HCl used:
moles of HCl = volume of HCl (L) x concentration of HCl (mol/L)
= 23.0 mL x 0.010 mol/L
= 0.230 mmol
Since 2 moles of HCl react with 1 mole of Mg(OH)2, we divide the number of moles of HCl by 2 to find the number of moles of Mg(OH)2:
moles of Mg(OH)2 = 0.230 mmol / 2
= 0.115 mmol
Next, we convert the moles of Mg(OH)2 to grams:
mass of Mg(OH)2 = moles of Mg(OH)2 x RFM of Mg(OH)2
= 0.115 mmol x 58 g/mol
= 6.67 g
Therefore, the mass of magnesium hydroxide in the anti-acid is 6.67 g.
b) To calculate the percentage of magnesium hydroxide in the anti-acid, divide the mass of magnesium hydroxide by the mass of the anti-acid and multiply by 100:
percentage of Mg(OH)2 = (mass of Mg(OH)2 / mass of anti-acid) x 100
= (6.67 g / 0.50 g) x 100
= 1334%
Therefore, the percentage of magnesium hydroxide in the anti-acid is 1334%.
c) To calculate the percentage of sodium nitrate that was converted to sodium nitrite, we need to determine the moles of sodium nitrate and sodium nitrite.
From the balanced equation, we can see that 2 moles of NaNO3 yield 2 moles of NaNO2.
First, calculate the number of moles of oxygen gas produced:
moles of O2 = mass of O2 (g) / molar mass of O2 (g/mol)
= 0.83 g / 32 g/mol
= 0.0259 mol
Since 2 moles of NaNO3 yield 1 mole of O2, we multiply the moles of O2 by 2 to find the moles of NaNO3:
moles of NaNO3 = 0.0259 mol x 2
= 0.0518 mol
Next, calculate the moles of NaNO2:
moles of NaNO2 = moles of NaNO3
= 0.0518 mol
Finally, calculate the percentage of sodium nitrate that was converted to sodium nitrite:
percentage conversion = (moles of NaNO2 / moles of NaNO3) x 100
= (0.0518 mol / 0.0518 mol) x 100
= 100%
Therefore, the percentage of sodium nitrate that was converted to sodium nitrite is 100%.