How many liters of oxygen gas (O2) are needed to produce 100 kJ of energy at STP? 2C6H6(I) + 1502(g) —> 12CO2(g)+6H2O(g)+3909.9 kJ

Note: Since no temperature and pressure conditions are stated, assume STP

2H2(g) + O2(g)



2H2O(g)

At STP, 2(22.4L) 22.4L

2. Dry ice is solid carbon dioxide. What volume of dry ice is produced at STP if 0.50 kg of

dry ice becomes carbon dioxide gas?

CO2 (s) 

CO2 (g)

At STP, 44.0 g 22.4 L

3. How many moles of steam are produced at 2.00 atm and 202.0°C by the complete

combustion of 12.50 L of methane gas?

CH4 + 2O2



CO2 + 2H2O

1 mol 2 mol

Note: Since temperature and pressure are stated, STP does not apply

Liter O2 = 40.0 L H2 x 22.4 L O2 = 20 Liters O2

44.8 L H2

Liter CO2 = 500 g CO2 x 22.4 L CO2 = 255 Liters CO2

44.0 g CO2

Given: P = 2.00 atm, T = 202.0 oC + 273 K/oC = 475 K, V CH4 = 12.50 L

First, use given info to find mole of CH4 from PV = nRT

n = PV/RT = 2.00 atm x 12.5 L = 0.641 mol CH4

0.0821 L.atm/mol.K x 475 K

Mol H2O = 0.641 mol CH4 x 2 mol H2O = 1.28 mol H2O

1 mol CH4

4. When nitrogen gas reacts with hydrogen gas, ammonia gas is formed. How many grams

of hydrogen gas are required to react completely with 3.5 L of nitrogen at STP?

N2 + 3H2



2NH3

At STP, 1(22.4L) 6.0g

5. Calculate the mass of water vapor produced when 6.00 L of propane gas undergo

complete combustion at a temperature of 250°C and a pressure of 1.00 atm.

C3H8 + 5O2



3CO2 + 4H2O

1 mol 4(18) g

Note: Since temperature and pressure are stated, STP does not apply

6. A gaseous compound is 30.6% nitrogen and 69.4% oxygen by mass. A 5.25-g sample of

the gas occupies a volume of 1.00 L and exerts a pressure of 1.26 atm at -4.0°C. What is

the molecular formula for the gas?

g H2 = 3.5 L N2 x 6.0 g H2_ = 0.94 g H2

22.4 L N2

Given: V C3H8 =6.00 L, T = 250 oC + 273 K/oC = 523 K, P = 1.00 atm

First, use given info to find mole of C3H8 from PV = nRT

n = PV/RT = 1.00 atm x 6.00 L = 0.1397 mol C3H8

0.0821 L.atm/mol.K x 523 K

Mass H2O = 0.1397 mol C3H8 x 72 g H2O_ = 10.1 g H2O

1mol C3H8

First, find the empirical formula

Element Mass % Number of mole Relative # mol (Divide by smaller # mol)

N 30.6 30.6 g__ = 2.1857 2.1857 = 1

14 g/mol 2.1857

O 69.4 69.4 g___ = 4.3375 4.3375 = 1.98 =2

16 g/mol 2.1857

Hence, empirical formula = NO2

Second, use given data to calculate the molar mass

Molar mass = massRT/PV

Molar mass = 5.25 g x 0.0821 L.atm/mol.K x 269 K = 92.0 g/mol

1.26 atm x 1.00 L

Remember that (empirical formula)x = molar mass

Therefore, (NO2)x = 92.0 Hence, 42x = 92.0; x

Explanation:

the answer is barium and copper bromide