Question

When tin comes in contact with the oxygen in the air, \(\operatorname{tin}(\mathrm{IV})\) oxide, \(\mathrm{SnO}_{2}\), is formed. $$ \mathrm{Sn}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SnO}_{2}(s) $$ A piece of tin foil, \(8.25 \mathrm{~cm} \times 21.5 \mathrm{~cm} \times 0.600 \mathrm{~mm}\) \(\left(d=7.28 \mathrm{~g} / \mathrm{cm}^{3}\right),\) is exposed to oxygen. (a) Assuming that all the tin has reacted, what is the mass of the oxidized tin foil? (b) Air is about \(21 \%\) oxygen by volume \((d=1.309 \mathrm{~g} / \mathrm{L}\) at \(\left.25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right) .\) How many liters of air are required to completely react with the tin foil?

Short answer

Answer: The mass of the oxidized tin foil is obtained by adding the mass of SnO₂ to the original mass of the tin foil. The volume of air required for the complete reaction with the tin foil can be calculated using the mass of O₂ required and the mass of O₂ in 1L of air.

Step-by-step solution

Calculate the mass of tin foil

To find out the mass of the tin foil, we first need to compute its volume and then multiply it by its density. The volume of tin foil is given by: Volume = Length × Width × Height The given dimensions are: Length = 8.25 cm Width = 21.5 cm Height = 0.600 mm = 0.0600 cm Density of tin = 7.28 g/cm³ Now calculate the volume and mass of the tin foil: Volume = 8.25 cm × 21.5 cm × 0.0600 cm Mass of tin foil = Volume × Density

Calculate the moles of tin foil

To perform stoichiometry calculations, we need to convert the mass of the tin foil into moles. We can do this by using the molar mass of tin: Molar Mass of Sn = 118.71 g/mol Now calculate the moles of tin foil: Moles of Sn = Mass of tin foil / Molar Mass of Sn

Calculate the moles of \(\mathrm{SnO}_{2}\) formed

The balanced chemical equation for the reaction is given: $$ \mathrm{Sn}(s) + \mathrm{o}_{2}(g) \longrightarrow \mathrm{SnO}_{2}(s) $$ From the balanced equation, we can see that 1 mole of Sn reacts with 1 mole of O₂ to form 1 mole of SnO₂. Therefore, the moles of SnO₂ formed are equal to the moles of Sn. Moles of SnO₂ = Moles of Sn

Calculate the mass of the oxidized tin foil

Now we need to convert the moles of SnO₂ formed to its mass. To do this, we can use the molar mass of SnO₂: Molar Mass of SnO₂ = 150.71 g/mol Calculate the mass of SnO₂ formed: Mass of SnO₂ = Moles of SnO₂ × Molar Mass of SnO₂ We will now find the mass of the oxidized tin foil by adding the mass of SnO₂ to the original mass of the tin foil: Mass of oxidized tin foil = Mass of tin foil + Mass of SnO₂

Calculate the moles of O₂ required

From the balanced equation, we know that 1 mole of Sn reacts with 1 mole of O₂. Therefore, the moles of O₂ required are equal to the moles of Sn: Moles of O₂ = Moles of Sn

Calculate the volume of air required

The percentage of oxygen in the air is given as 21%. We can now calculate the mass of O₂ present in the air using the given information about the density of air at given conditions (T = 25°C, P = 1 atm): Density of air = 1.309 g/L Percentage of oxygen in air = 21% Now calculate the mass of O₂ in 1L of air: Mass of O₂ in 1L of air = 1.309 g/L × 0.21 We can now calculate the volume of air required for the complete reaction with the tin foil. First, convert the moles of O₂ to mass using the molar mass of O₂: Molar Mass of O₂ = 32.00 g/mol Mass of O₂ required = Moles of O₂ × Molar Mass of O₂ Now calculate the volume of air required for the reaction: Volume of air required = Mass of O₂ required / (Mass of O₂ in 1 L of air) The obtained value will be the volume of air in liters required to completely react with the tin foil.