Question

Tin(IV) chloride is used as an external coating on glass to toughen glass containers. It is prepared by reacting tin with chlorine gas. $$ \mathrm{Sn}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SnCl}_{4}(l) $$ The process requires a \(25.0 \%\) excess of tin and has a yield of \(73.7 \% .\) Calculate the mass of tin and the volume of chlorine gas \(\left(d=2.898 \mathrm{~g} / \mathrm{L}\right.\) at \(\left.25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)\) needed to produce $$ 0.500 \mathrm{~L} \text { of } \mathrm{SnCl}_{4}(d=2.226 \mathrm{~g} / \mathrm{mL}) $$

Short answer

The mass of tin required is 634.67 g, and the volume of chlorine gas required is 70.98 L.

Step-by-step solution

Convert volume of SnCl4 to mass

To find the mass of SnCl4 produced, we first need to convert its volume (0.500 L) into mass using the given density (2.226 g/mL): $$ Mass \: of \: SnCl_4 = Density (g/ml) \times Volume (L) \times 1000\: (Conversion \: Factor) $$ $$ Mass \: of \: SnCl_4 = 2.226 \frac{g}{mL} \times 0.500 \: L \times 1000 \frac{mL}{L} = 1113 \: g $$

Stoichiometry to calculate mass of tin and volume of chlorine gas

Using stoichiometry, we can determine the raw mass of tin required for the SnCl4 production as well as the volume of chlorine gas required. First, we need to calculate the molar mass of Sn, Cl2, and SnCl4: - Molar mass of Sn: 118.71 g/mol - Molar mass of Cl2: 2 * 35.45 g/mol = 70.9 g/mol - Molar mass of SnCl4: Sn + 4*Cl = 118.71 g/mol + 4*35.45 g/mol = 260.56 g/mol Next, we can convert the mass of SnCl4 to moles and find the moles of tin and chlorine gas required using the stoichiometry ratios from the balanced chemical equation: $$ Moles \: of \: SnCl_4 = \frac{Mass \: of \: SnCl_4}{Molar \: Mass \: of \: SnCl_4} $$ $$ Moles \: of \: SnCl_4 = \frac{1113\: g}{260.56 \: g/mol} = 4.276 \: mol $$ Now we can use the stoichiometry ratios: $$ Moles \: of \: Sn = Moles \: of \: SnCl_4 = 4.276 \: mol $$ $$ Moles \: of \: Cl_2 = \frac{1}{2} Moles \: of \: SnCl_4 = \frac{1}{2} \times 4.276 \: mol = 2.138 \: mol $$ Then, we can convert moles of Sn and Cl2 to masses using their respective molar masses: $$ Mass \: of \: Sn = moles \: of \: Sn \times Molar \: Mass \: of \: Sn = 4.276 \: mol \times 118.71 \: g/mol = 507.74 \: g $$ $$ Mass \: of \: Cl_2 = moles \: of \: Cl_2 \times Molar \: Mass \: of \: Cl_2 = 2.138 \: mol \times 70.9 \: g/mol = 151.66 \: g $$

Adjust mass of tin for excess required

We need to account for the 25.0% excess of tin to calculate the adjusted mass of tin required: $$ Adjusted \: Mass \: of \: Tin = Mass \: of \: Tin \times (1 + Excess \: Percentage) = 507.74 \: g \times 1.25 = 634.67 \: g $$

Calculate the volume of chlorine gas

To find the volume of chlorine gas at the given conditions, we can first convert the mass of chlorine gas to volume using the given density (2.898 g/L): $$ Volume \: of \: Cl_2 = \frac{Mass \: of \: Cl_2}{Density \: of \: Cl_2} = \frac{151.66 \: g}{2.898 \: g/L} = 52.31 \: L $$ Since the percentage yield is 73.7%, we can find the required volume of chlorine gas (necessary to produce the desired amount of SnCl4) by dividing the calculated volume by the yield: $$ Required \: volume \: of \: Cl_2 = \frac{Volume \: of \: Cl_2}{Yield} = \frac{52.31 \: L}{0.737} = 70.98 \: L $$ The mass of tin required is 634.67 g and the volume of chlorine gas required is 70.98 L to produce 0.500 L of SnCl4 under the given conditions.