SHOW WORK for points. 1. A 1.20 g sample of dry ice is added to a 755 mL flask c

Question

SHOW WORK for points.

1. A 1.20 g sample of dry ice is added to a 755 mL flask containing N2 gas at 25.0 °C and 725 mmHg. If all of the dry ice sublimes (converts from a solid to a gas) and the temperature in the flask then returns to 25.0 °C, what is the total pressure in the flask?

2. A mixture of gasses contains 1.25 g N2 and 0.85 g O2 in a 1.55 L flask at 18 °C. Calculate the mole fraction and partial pressure of each component in the gas mixture.

3. CH3OH is synthesized via the following reaction: CO (g) + 2H2 (g) CH3OH (g) . If the reaction occurs at 748 mmHg and 25 °C , what volume of H2 gas (in L) is necessary to synthesize 25.8 g CH3OH? What volume (L) of CO gas is required?

4. Hydrogen gas can be formed by the reaction of methane and water according to the following equation: CH4 (g) + H2O (g) CO (g) + 3H2 (g) . In one reaction, 25.5 L of methane gas (at P = 732 Torr, T = 25 °C) is mixed with 22.8 L of water vapor (at P = 702 Torr, T = 125 °C). The reaction produces 26.2 L hydrogen gas at STP. What is the percent yield of the reaction?

Explanation / Answer

you have posted so many questions i will answer some of them

1) we have to use idal gas equation

PV = nRT i hope you know what these letters will indicate

p = 725 mmHg = 0.95 atm
V = 755 mL = 0.755 Liter
R = 0.0821 (l x atm) / (mol x K)
T = 25°C = = 298.16 K

plug in all the values in above equation then you will get n = 0.029 mol N2

no of moles of CO2 = 1.20 / 44 = 0.027 mol

total no of moles in the flask (because CO2 get sublimated that will also in gas form

0.027 + 0.029 = 0.056 mol

now we know the moles of gas mixture

again use the PV = nRT to get the pressure

n = = 0.056 mol
V = 0.755 Liter
R = 0.0821 (l * atm) / (mol * K)
T = 298.16 K

plug the values

then p = 1.82 atm

2)

first we need to find out the mole of these two gases using wt/mwt

1.25 g / 28 g = 0.045 mol N2
0.85 g / 32 g = 0.027 mol O2

total moles = 0.045 + 0.027 = 0.072 mol

now the mole fraction

N2 mol fraction 0.045/ 0.072 = 0.625 mol

O2 mol fraction 0.027/ 0.072 = 0.375 mol

partial pressures can be find out again using PV =nRT

partial pressure of N2 =(0.045 mol)(0.0821)(291 K) / 1.55 L = 0.6936

partial pressure of O2 (0.027 mol)(.0821)(291 K) / 1.55 = 0.4161

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