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)=Sn2+(aq)+H2(g)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×ΔVA = 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×V(H2)A = p \times V(H2)

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

p×V=n×R×Tp \times V = n \times R \times T

V=n×R×TpV = \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)=mMn(Sn) = \frac{m}{M}

n(Sn) = 0.42 mole

V=0.42×8.31×2981.013×105=0.01m3V = \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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