1. A cylinder contains 28.0 L of oxygen gas at a pressure of 1.6 atm and a tenpr

Question

1. A cylinder contains 28.0 L of oxygen gas at a pressure of 1.6 atm and a tenprerature of 288 K. How many moles of gas are in the cylinder?
2. A mixture of helium, nitrogen and oxygen has atoral pressure of 762 mm Hg. The partial pressures of helium and nitrogen are 239 mm Hg and 203 mm Hg. What is the partial pressure of oxygen in the mixture?
3.The hydrogen gas formed in a chemical reaction is collected over water at 30 C at a total pressure of 738 mm Hg. What is the partial pressure of the hydrogen gas collected in this way?
4. Calculate the volume pf each gas sample at STP. A) 21.6 mol Cl2 B) 4.3 mol nitrogen C) 2.4 mol helium D) 29 mol CH4
5. Consider the chemical equation: C(s) + H2O(g) --> CO (g) + H2 (g) How many liters of hydrogen gas are formed from the complete reaction of 1.07 mol of C? Assume that the hydrogen gas is collected at a pressure of 1.0 atm and temperature of 313 K
6. Nitrogen reacts with powdered aluminum according to the reaction: 2 Al(s) + N2 (g) --> 2AlN(s) How many liters of N2 gas, measured at 891 torr and 92 C are required to completely react with 20.3g of Al?
1. A cylinder contains 28.0 L of oxygen gas at a pressure of 1.6 atm and a tenprerature of 288 K. How many moles of gas are in the cylinder?
2. A mixture of helium, nitrogen and oxygen has atoral pressure of 762 mm Hg. The partial pressures of helium and nitrogen are 239 mm Hg and 203 mm Hg. What is the partial pressure of oxygen in the mixture?
3.The hydrogen gas formed in a chemical reaction is collected over water at 30 C at a total pressure of 738 mm Hg. What is the partial pressure of the hydrogen gas collected in this way?
4. Calculate the volume pf each gas sample at STP. A) 21.6 mol Cl2 B) 4.3 mol nitrogen C) 2.4 mol helium D) 29 mol CH4
5. Consider the chemical equation: C(s) + H2O(g) --> CO (g) + H2 (g) How many liters of hydrogen gas are formed from the complete reaction of 1.07 mol of C? Assume that the hydrogen gas is collected at a pressure of 1.0 atm and temperature of 313 K
6. Nitrogen reacts with powdered aluminum according to the reaction: 2 Al(s) + N2 (g) --> 2AlN(s) How many liters of N2 gas, measured at 891 torr and 92 C are required to completely react with 20.3g of Al?

2. A mixture of helium, nitrogen and oxygen has atoral pressure of 762 mm Hg. The partial pressures of helium and nitrogen are 239 mm Hg and 203 mm Hg. What is the partial pressure of oxygen in the mixture?
3.The hydrogen gas formed in a chemical reaction is collected over water at 30 C at a total pressure of 738 mm Hg. What is the partial pressure of the hydrogen gas collected in this way?
4. Calculate the volume pf each gas sample at STP. A) 21.6 mol Cl2 B) 4.3 mol nitrogen C) 2.4 mol helium D) 29 mol CH4
5. Consider the chemical equation: C(s) + H2O(g) --> CO (g) + H2 (g) How many liters of hydrogen gas are formed from the complete reaction of 1.07 mol of C? Assume that the hydrogen gas is collected at a pressure of 1.0 atm and temperature of 313 K
6. Nitrogen reacts with powdered aluminum according to the reaction: 2 Al(s) + N2 (g) --> 2AlN(s) How many liters of N2 gas, measured at 891 torr and 92 C are required to completely react with 20.3g of Al?

Explanation / Answer

1)

for gases

PV = nRT

so

1.6 x 28= n x 0.0821 x 288

n = 1.895

so

moles of gas is 1.895

2)

we know that

total pressure = pHe + pN2 + p02

so

762 = 239 + 203 + p02

p02 = 320

so

partial pressure of oxygen is 320 mm Hg

3)

pressure of hydrogen gas= total pressure - vapor pressure of water at 30 C

pressure of hydrogen gas = 738 - 31.8

pressure of hydrogen gas = 706.2 mm Hg

4)

we know that

for any gas at STP

volume of gas = moles x 22.4 L

so

a) volume of Cl2 = 21.6 x 22.4 = 483.84 L

b) volume of nitrogen = 4.3 x 22.4 = 96.32

c) volume of helium = 2.4 x 22.4 = 53.76 L

d) volume of CH4 = 29 x 22.4 = 649.6 L

5)

C(s) + H20 ---> CO + H2

we can see that

moles of H2 formed = moles of C reacted = 1.07

now

PV = nRT

1 x V = 1.07 x 0.0821 x 313

V = 27.5 L

so

27.5 L of hydrogen gas is formed

6)

we know that

moles = mass / molar mass

so

moles of Al = 20.3/27

moles of Al = 0.7519

now

2Al + N2 ---> 2 AlN

we can see that

moles of N2 required = 0.5 x moles of Al

moles of N2 required = 0.5 x 0.7519 = 0.37595

now

PV = nRT

(891/760) x V = 0.37595 x 0.0821 x 365

V= 9.61

so

9.61 L of nitrogen gas is required

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