A0.5895-g sample of impure magnesium hydroxide is dissolved in 100.0 ml of 0.2050 m hcl solution. the excess acid then needs 19.85 ml of 0.1020 m naoh for neutralization. calculate the percent by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the hcl solution. (can some show all the steps to this with the answer? i’m completely lost).

91.38 %

Explanation:

First we use the following formula to calculate the number of moles of HCl and NaOH contained in the solutions:

molar concentration = number of moles / solution volume

number of moles = molar concentration × solution volume

number of moles of HCl = 0.2050 × 100 = 20.5 mmoles

number of moles of NaOH = 0.1020 × 19.85 = 2.02 mmoles

reaction between NaOH and HCl:

NaOH + HCl → NaCl + H₂O

If 2.02 mmoles of NaOH were used for neutralization of HCl left from the reaction, which are 2.02 mmoles of HCl, it means that the HCl reacted with Mg(OH)₂:

20.5 - 2.02 = 18.48 mmoles of HCl

2 HCl + Mg(OH)₂ = MgCl₂ + 2 H₂O

From the chemical reaction we devise the following reasoning:

if       2 mmoles of HCl react with 1 mmole of Mg(OH)₂

then  18.48 mmoles HCl react with X mmoles of Mg(OH)₂

X = (18.48 × 1) / 2 = 9.24 mmoles of Mg(OH)₂

now the mass of Mg(OH)₂:

mass = number of moles × molecular weight

mass of Mg(OH)₂ = 9.24 × 58.3 = 538.7 mg = 0.5387 g

Now the percent by mass of magnesium hydroxide Mg(OH)₂ in the impure sample (purity):

purity = (mass of pure Mg(OH)₂ / mass of impure sample) × 100

purity = (0.5387 / 0.5895) × 100

purity = 91.38 %

1 a

2 d

3 a

4 c

explanation: