Answer to Question #93572 in Chemistry for Nisa

Question

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]

Answer 1

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