Answer to Question #93572 in Chemistry for Nisa

Question #93572
ulate the work done when 50.0 g of tin dissolvesin excess acid at 1.00 atm and 25°C: Sn(s) +2H+(aq) Sn2+(aq) +H2(g) [Molar mass of tin 118.7 g]
1
Expert's answer
2019-09-02T08:10:48-0400

Solution.

"Sn(s) +2H^+(aq) = Sn^{2+}(aq) +H2(g)"

Since, according to the condition, the process occurs at constant pressure, and, therefore, the work will be equal to:

"A = p \\times \\Delta V"

In this case, the difference between the final and initial volumes of the system is essentially equal to the volume of gaseous H2 resulting from the reaction.

"A = p \\times V(H2)"

We write the Clapeyron-Mendeleev equation (assuming that an ideal gas is formed upon dissolution of tin):

"p \\times V = n \\times R \\times T"

"V = \\frac{n \\times R \\times T}{p}"

n(Sn) = n(H2), since the amount of tin substance, according to the reaction equation, is equal to the amount of hydrogen substance.

"n(Sn) = \\frac{m}{M}"

n(Sn) = 0.42 mole

"V = \\frac{0.42 \\times 8.31 \\times 298}{1.013 \\times 10^5} = 0.01 m^3"

A = 1013 J

Answer:

A = 1013 J


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