Question

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
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b. the flask with 0.6 g mg
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c. the flask with 1.2 g mg
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Explanation:

Step1: Calculate moles of Mg

The molar mass of Mg is \(M = 24.3\space g/mol\).
For \(m = 0.3\space g\) Mg: \(n_{0.3}=\frac{m}{M}=\frac{0.3}{24.3}\space mol\)
For \(m = 0.6\space g\) Mg: \(n_{0.6}=\frac{m}{M}=\frac{0.6}{24.3}\space mol\)
For \(m = 1.2\space g\) Mg: \(n_{1.2}=\frac{m}{M}=\frac{1.2}{24.3}\space mol\)

The reaction of Mg with an acid (e.g., \(Mg + 2H^{+}=Mg^{2 +}+H_{2}\uparrow\)) has a 1:1 mole ratio between Mg and \(H_{2}\).

Step2: Normalize the moles of \(H_{2}\)

For part a:
The normalization factor is \(\frac{n_{0.3}}{n_{0.6}}\).
Since \(n_{0.3}=\frac{0.3}{24.3}\) and \(n_{0.6}=\frac{0.6}{24.3}\), then \(\frac{n_{0.3}}{n_{0.6}}=\frac{\frac{0.3}{24.3}}{\frac{0.6}{24.3}} = 0.5\)

For part c:
The normalization factor is \(\frac{n_{1.2}}{n_{0.6}}\)
Since \(n_{1.2}=\frac{1.2}{24.3}\) and \(n_{0.6}=\frac{0.6}{24.3}\), then \(\frac{n_{1.2}}{n_{0.6}}=\frac{\frac{1.2}{24.3}}{\frac{0.6}{24.3}}=2\)

Answer:

a. \(0.5\)
b. \(1\)
c. \(2\)