1. Tin and oxygen gas are reacted together to form Tin (IV) oxide, a solid. If a piece of tin foil, 8.25cm x 21.5cm x 0.60mm (density = 7.28g/cm3) is exposed to oxygen and all the tin reacts, what is the mass of the oxidized tin foil? (98.4g)
2. From the information in problem 1, if air is about 21% oxygen by volume (d=1.309g/L) at 25C, atm). How many liters of air are required to completely react ith the tin foil? (76L)
3. Oxygen masks for producing O2 in emergency situations contain potassium superoxide, KO2. It reacts with CO2 and H2O in exhaled air to produce oxygen and potassium hydrogen carbonate. If a person wearing such a mask exhales 0.70g carbon dioxide perminute, how many grams of KO2 are consumed in 25 minutes? (28g)
Can you start with finding the weight of the tin foil?
Tin (IV) means that the resulting compound is Sn++++, or equivalent meaning:
Sn + 2O2 = SnO4
With the ATOMIC mass of tin=118.17, and 16 for O2, you should be able to complete the calculations.
Sn(IV) oxide is SnO2.
The correct equation is
Sn + O2 ==> SnO2.
Oops, thanks Dr. Bob.
Can someone explain how to do this?
To find the answers to these questions, we can use stoichiometry and dimensional analysis.
1. To find the mass of the oxidized tin foil, we need to calculate the mass of tin (IV) oxide formed. First, we determine the volume of the tin foil:
Volume = length x width x height
Volume = 8.25 cm x 21.5 cm x 0.60 mm
Note: Since density is given in g/cm3, the dimensions need to be in the same units, so we convert the height from mm to cm:
Volume = 8.25 cm x 21.5 cm x 0.06 cm
Next, we multiply the volume by the density to find the mass:
Mass = Volume x Density
Mass = (8.25 cm x 21.5 cm x 0.06 cm) x 7.28 g/cm3
Mass = 98.4 g
2. To calculate the volume of air required to react with the tin foil, we need to determine the amount of oxygen gas involved in the reaction. Since air is about 21% oxygen by volume, we can calculate the volume of oxygen gas using the stoichiometry of the reaction. We can assume that the reaction consumes all the tin and the amount of oxygen gas consumed is stoichiometrically equivalent to the amount of tin.
To convert volume to moles of gas, we can use the ideal gas law:
PV = nRT
where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. At 25°C, the pressure is 1 atm. Rearranging the equation, we can solve for n:
n = PV / RT
where P is in atm, V is in liters, R is 0.0821 atm L/mol K, and T is in Kelvin. Since we are given the density of air, we can use it to convert the volume of air to moles of oxygen gas using the molar volume at STP (standard temperature and pressure):
Molar volume at STP = 1 mole / 22.4 liters
Thus, we can calculate the number of moles of oxygen gas:
Moles of oxygen gas = (Volume of air in liters) x (Oxygen percentage in decimal) x (1 mole / 22.4 liters)
After finding the number of moles of oxygen gas, we can convert it to liters by using the molar volume at STP to find the volume of air required.
3. To determine the grams of KO2 consumed in 25 minutes, we can use stoichiometry and the molar mass of carbon dioxide (CO2). From the given information, we know that 0.70 g of carbon dioxide is exhaled per minute.
First, we need to calculate the number of moles of carbon dioxide exhaled in 25 minutes:
Moles of CO2 = (Mass of CO2 exhaled per minute) x (1 mole / Molar mass of CO2) x (Time in minutes)
After calculating the number of moles of carbon dioxide, we can use the stoichiometry of the reaction to determine the moles of KO2 consumed. From the balanced equation, we know that the stoichiometric ratio between CO2 and KO2 is 1:1.
Finally, we can convert the moles of KO2 to grams by using the molar mass of KO2.