Question

A special gas mixture is used in bacterial growth chambers which contains 5.00% by mass of CO2 and 95.0% O2. If the mole fraction of CO2 and O2 in the mixture is 0.0998 and 0.2340, respectively. What is the partial pressure in atmospheres of O2 and CO2 gases at a total pressure of 0.6880 atm?

Answer 1

First, we can use the mole fractions to calculate the number of moles of each gas present in the mixture. Let's assume we have 100 grams of the mixture, then:

- Moles of CO2 = 0.05 x 100 / 44.01 = 0.1136 moles
- Moles of O2 = 0.95 x 100 / 32.00 = 2.9688 moles

Next, we can use the total pressure and mole fractions to calculate the partial pressures of each gas. The partial pressure of a gas is equal to its mole fraction times the total pressure. So:

- Partial pressure of CO2 = 0.0998 x 0.6880 atm = 0.0685 atm
- Partial pressure of O2 = 0.2340 x 0.6880 atm = 0.1608 atm

Therefore, the partial pressure of CO2 in the mixture is 0.0685 atm and the partial pressure of O2 is 0.1608 atm.

Answer 2

If the theoretical yield of PbCl2 is 16.547g from the reaction of 7.000g of NaCl with 50.400g of Pb(NO3)2. Calculate the % yield of PbCl2 if 12.788g of PbCl2 was actually obtained? (Atomic mass: Na-22.990 g/mol; Cl-35.453g/mol; 0-15.999g/mol; Pb-207.200 g/mol; N-14.007).

Answer 3

First, let's balance the equation:

2NaCl + Pb(NO3)2 → 2NaNO3 + PbCl2

From the balanced equation, the stoichiometry of the reaction is 2 moles of NaCl to 1 mole of Pb(NO3)2 to 1 mole of PbCl2.

We can calculate the number of moles of Pb(NO3)2 used in the reaction as follows:

50.400 g Pb(NO3)2 × 1 mol Pb(NO3)2 / 331.209 g Pb(NO3)2 = 0.1521 mol Pb(NO3)2

Therefore, the number of moles of PbCl2 that would be produced if the reaction went to completion is also 0.1521 mol (1:1 stoichiometry with Pb(NO3)2). The theoretical yield of PbCl2 can be calculated from the molar mass:

16.547 g PbCl2 × 1 mol PbCl2 / 278.104 g PbCl2 = 0.0595 mol PbCl2

Now we can calculate the percent yield:

Percent yield = actual yield / theoretical yield × 100%
= 12.788 g / 16.547 g × 100%
= 77.29%

Therefore, the percent yield of PbCl2 is 77.29%.