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Solved: Calculate the work done when 50.0 g of tin
Chapter 6, Problem 19P(choose chapter or problem)
Calculate the work done when 50.0 g of tin dissolves in excess acid at 1.00 atm and 25°C:
\(\mathrm{Sn}(\mathrm{s})+2\mathrm{H}^+(\mathrm{aq})\rightarrow\mathrm{Sn}^{2+}(\mathrm{aq})+\mathrm{H}_2(\mathrm{g})\)
Assume ideal gas behavior.
Questions & Answers
QUESTION:
Calculate the work done when 50.0 g of tin dissolves in excess acid at 1.00 atm and 25°C:
\(\mathrm{Sn}(\mathrm{s})+2\mathrm{H}^+(\mathrm{aq})\rightarrow\mathrm{Sn}^{2+}(\mathrm{aq})+\mathrm{H}_2(\mathrm{g})\)
Assume ideal gas behavior.
ANSWER:Step 1 of 2
The goal of the problem is to calculate the work done
Given equation:
Given:
mass = 50.0 g
P = 1.00 atm
T = 25°C = 25 + 273 = 298 K
First, let’s calculate the number of moles of hydrogen gas formed in the reaction:
Molar mass of Sn (Tin) = 118.71 g/mol
We see from the balance equation that, there is 1 mole of for 1 mole of Sn. Therefore, the number of moles will be:
= 50.0 g 
= 0.421 moles
Now, let’s calculate the volume occupied by the hydrogen gas using the ideal gas equation:
PV = nRT
V =
=
= 10.3 L
Hence, the volume of hydrogen gas released is 10.3 L