Question

100.0 ml of hcl required 5.831 g mg(oh)₂ to titrate. the molar mass of mg(oh)₂ is 58.31 g. what is the concentration of the hcl? the neutralization equation is 2 hcl + mg(oh)₂ → mgcl₂ + 2 h₂o there is enough information to calculate the moles of mg(oh)₂ 1 mol mg(oh)₂ (5.831 g mg(oh)₂ (----------------) = 2 mol mg(oh)₂ 3 g mg(oh)₂ then use stoichiometry to convert moles of mg(oh)₂ into moles of hcl 4 mol hcl (5 mol mg(oh)₂)(----------------) = 6 mol hcl 7 mol mg(oh)₂ we now can calculate the concentration of hcl since we have both the moles of hcl and the volume of hcl 8 mol hcl concentration = ------------ = 9 mol/l 10 l hcl a. acid b. base c. hcl d. naoh e. nacl f. h₂o g. 25.00 h. 0.02500 i. 0.2000 j. 1.000 k. 0.005000 l. 100.0 m. 0.1000 n. 0.05000 o. 2.000 p. 0.01000 q. 0.001000 r. 58.31 s. 5.831

Explanation:

Step1: Calculate moles of \(Mg(OH)_2\)

The formula for calculating moles \(n=\frac{m}{M}\), where \(m\) is mass and \(M\) is molar mass.
\(n_{Mg(OH)_2}=\frac{5.831\space g}{58.31\space g/mol}\)

Step2: Use stoichiometry

From the equation \(2HCl + Mg(OH)_2
ightarrow MgCl_2 + 2H_2O\), the mole ratio of \(HCl\) to \(Mg(OH)_2\) is \(2:1\).
If \(n_{Mg(OH)_2} = 0.1\space mol\) (from step 1: \(\frac{5.831}{58.31}=0.1\)), then \(n_{HCl}=2\times n_{Mg(OH)_2}\)

Step3: Calculate concentration of \(HCl\)

Concentration \(c=\frac{n}{V}\), where \(V = 100.0\space mL=0.1\space L\)

Answer:

1. \(1\)
2. \(0.1\) (since \(\frac{5.831}{58.31} = 0.1\))
3. \(58.31\)
4. \(2\)
5. \(0.1\)
6. \(0.2\) (since \(0.1\times2\))
7. \(1\)
8. \(0.2\)
9. \(2.000\) (since \(\frac{0.2\space mol}{0.1\space L}=2.000\space mol/L\))
10. \(0.1\) (since \(100.0\space mL = 0.1\space L\))