Question

25 ml of hydroiodic acid generates 1.20 l of h₂ gas at stp conditions when combined with excess zinc. what is the concentration (mol/l) of the hydroiodic acid? zn + 2hi → zni₂ + h₂ ? m hi hint: first use molar volume to solve for h₂; 1 mole of any gas at stp = 22.4 l.

Explanation:

Step1: Calculate moles of \( H_2 \)

At STP, 1 mole of gas occupies 22.4 L. So moles of \( H_2 \) = volume of \( H_2 \) / molar volume.
\( n_{H_2} = \frac{1.20\ L}{22.4\ L/mol} \)

Step2: Relate moles of \( H_2 \) to moles of \( HI \)

From the reaction \( Zn + 2HI
ightarrow ZnI_2 + H_2 \), the mole ratio of \( HI \) to \( H_2 \) is 2:1. So moles of \( HI \) = 2 × moles of \( H_2 \).
\( n_{HI} = 2\times\frac{1.20\ L}{22.4\ L/mol} \)

Step3: Calculate concentration of \( HI \)

Concentration (M) = moles / volume (in L). Volume of \( HI \) is 25 mL = 0.025 L.
\( M_{HI} = \frac{2\times\frac{1.20\ L}{22.4\ L/mol}}{0.025\ L} \)
First, calculate \( 2\times\frac{1.20}{22.4} = \frac{2.40}{22.4} \approx 0.1071\ mol \)
Then, \( M_{HI} = \frac{0.1071\ mol}{0.025\ L} \approx 4.28\ mol/L \)

Answer:

\( \approx 4.28\ M \) (or more precisely, using exact calculations: \( \frac{2\times1.20}{22.4\times0.025}=\frac{2.4}{0.56}\approx4.2857\ M \))