Question

b.4 calculate the number of moles of h₂ that are expected to form in each flask based on the number of moles of mg that will react and the equation for each reaction. (1 mole of h₂ is produced for every mole of mg reacted.)
a. the flask with 0.3 g mg

b. the flask with 0.6 g mg

c. the flask with 1.2 g mg

b.5 normalize the number of moles of h₂ that are produced by dividing the number of moles of h₂ for each reaction by the number of moles for the reaction of 0.6 g mg.
a. the flask with 0.3 g mg

b. the flask with 0.6 g mg

c. the flask with 1.2 g mg

questions and problems
q.4 how do the ratios in calculation b.5 compare with the observations of the volume of gas in each balloon in a.2, a.3 and a.4? were these ratios in b.5 close to the experimental observations? comment if there were any significant differences.

Explanation:

Step1: Calculate moles of Mg

The molar mass of Mg is \(M = 24.305\space g/mol\). The formula for moles \(n=\frac{m}{M}\), where \(m\) is mass.
For part a: \(n_{Mg}=\frac{0.3\space g}{24.305\space g/mol}\approx0.0123\space mol\)
For part b: \(n_{Mg}=\frac{0.6\space g}{24.305\space g/mol}\approx0.0247\space mol\)
For part c: \(n_{Mg}=\frac{1.2\space g}{24.305\space g/mol}\approx0.0494\space mol\)

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

Given that 1 mole of \(H_2\) is produced for every mole of Mg reacted. So \(n_{H_2}=n_{Mg}\)
For part a: \(n_{H_2}\approx0.0123\space mol\)
For part b: \(n_{H_2}\approx0.0247\space mol\)
For part c: \(n_{H_2}\approx0.0494\space mol\)

Step3: Normalize for B.5

For part a: \(\frac{0.0123}{0.0247}\approx0.5\)
For part c: \(\frac{0.0494}{0.0247} = 2\)

Answer:

B.4:
a. \(0.0123\space mol\)
b. \(0.0247\space mol\)
c. \(0.0494\space mol\)

B.5:
a. \(0.5\)
c. \(2\)

Q.4:
The ratios in B.5 (0.5, 1, 2) should be directly proportional to the volume of gas in the balloons (assuming ideal gas behavior \(V = nRT/P\), where \(R\), \(T\), \(P\) are constant). If the experimental observations (from A.2, A.3, A.4) show volumes in a 1:2:4 ratio (for example, if 0.3g, 0.6g, 1.2g Mg cases), there might be a discrepancy. But if the volumes are in a 1:2:4 (since moles ratios from B.5 for 0.3g:0.6g:1.2g are 0.5:1:2 and if volume is proportional to moles \(V\propto n\)), the ratios should be close. Any significant differences could be due to non - ideal gas behavior (e.g., gas not at ideal conditions \(T\), \(P\) not constant), leaks in the system, or inaccuracies in measuring mass of Mg or volume of gas.