Question

general chemistry 4th edition
mcquarrie · rock · gallogly
university science books
presented by macmillan learning
chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, hcl(aq), as described
by the chemical equation
mno₂(s) + 4 hcl(aq) → mncl₂(aq) + 2 h₂o(l) + cl₂(g)
how much mno₂(s) should be added to excess hcl(aq) to obtain 225 ml cl₂(g) at 25 °c and 735 torr?

Explanation:

Step1: Convert units

Convert pressure \(P = 735\ Torr\) to \(atm\): \(P=\frac{735}{760}\ atm\approx0.967\ atm\).
Convert volume \(V = 225\ mL\) to \(L\): \(V = 0.225\ L\).
Convert temperature \(T=25^{\circ}C\) to \(K\): \(T=(25 + 273)\ K=298\ K\).

Step2: Use ideal gas law \(PV = nRT\) to find \(n(Cl_2)\)

\(n(Cl_2)=\frac{PV}{RT}\), where \(R = 0.0821\ L\cdot atm\cdot K^{-1}\cdot mol^{-1}\).
\(n(Cl_2)=\frac{0.967\times0.225}{0.0821\times298}\)

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Step3: Use stoichiometry

From the balanced equation \(MnO_2(s)+4HCl(aq)\to MnCl_2(aq)+2H_2O(l)+Cl_2(g)\), the mole ratio \(n(MnO_2):n(Cl_2)=1:1\). So \(n(MnO_2)=n(Cl_2) = 0.0089\ mol\).

Step4: Calculate mass of \(MnO_2\)

The molar mass of \(MnO_2\) is \(M = 86.94\ g/mol\).
\(m(MnO_2)=n(MnO_2)\times M\)
\(m(MnO_2)=0.0089\ mol\times86.94\ g/mol\approx0.774\ g\)

Answer:

\(0.774\ g\)