Question

b. calculations
b.1 calculate the number of moles of hcl added to each flask (a.1 × 0.05l)

b.2 calculate the number of moles of magnesium ribbon added to each flask.
a. 0.3 g/atomic weight of mg (24.3 g/mole)
b. 0.6 g/ atomic weight of mg (24.3 g/mole)
c. 1.2 g/ atomic weight of mg (24.3 g/mole)

Explanation:

Step1: Calculate moles of HCl

The formula for moles \(n = c\times V\). Assuming \(c = 1\space mol/L\) (since \(A.1\) is likely \(1\space mol/L\) concentration), \(V=0.05\space L\).
\(n = 1\times0.05\)

Step2: Calculate moles of Mg (a)

Using formula \(n=\frac{m}{M}\), where \(m = 0.3\space g\), \(M = 24.3\space g/mol\)
\(n=\frac{0.3}{24.3}\)

Step3: Calculate moles of Mg (b)

Using formula \(n=\frac{m}{M}\), where \(m = 0.6\space g\), \(M = 24.3\space g/mol\)
\(n=\frac{0.6}{24.3}\)

Step4: Calculate moles of Mg (c)

Using formula \(n=\frac{m}{M}\), where \(m = 1.2\space g\), \(M = 24.3\space g/mol\)
\(n=\frac{1.2}{24.3}\)

Answer:

B.1: \(0.05\space mol\)
B.2 a: \(\approx0.0123\space mol\)
B.2 b: \(\approx0.0247\space mol\)
B.2 c: \(\approx0.0494\space mol\)