Question

1. Hydrogen produced in the following reaction is collected over water at 23oC and 742 Torr: 2Al + 6HCl --> 2AlCl3 + 3H2 What volume (mL) of the gas will be collected in the reaction of 1.50 g Al with excess HCl?

2. Sample of O2 gas has a volume of 50.0 L at a pressure of 750. mmHg. What is the final volume, in liters, of the gas at 2.0 atm?

3. What is the mass, in grams, of 475 mL of nitrogen gas at STP?

4. A 25.0 mL bubble is released from a diver's air tank at a pressure of 4.00 atm and a pressure of 11oC. What is the volume, in mL, of the bubble when it reaches the ocean surface, where the pressure is 1.00 atm and the temperature is 18oC?

5. You vaporize a liquid substance at 100 oC and 755 mmHg. The volume of 0.548 g of vapor is 237 mL. What is the molecular mass of the substance?

6. Given the following chemical equation: 4NH3 + 5O2 --> 4NO + 6H2O What volume (mL) of oxygen at 35 oC and 2.15 atm is needed to produce 100.0 g of NO?

Answer 1

1)

Molar mass of Al = 26.98 g/mol

mass of Al = 1.5 g

mol of Al = (mass)/(molar mass)

= 1.5/26.98

= 0.0556 mol

we have the Balanced chemical equation as:

2Al + 6HCl --> 2AlCl3 + 3H2

From balanced chemical reaction, we see that

when 2 mol of Al reacts, 3 mol of H2 is formed

mol of H2 formed = (3/2)* moles of Al

= (3/2)*0.0556

= 0.0834 mol

we have:

P = 742.0 torr

= (742.0/760) atm

= 0.9763 atm

n = 0.0834 mol

T = 23.0 oC

= (23.0+273) K

= 296 K

we have below equation to be used:

P * V = n*R*T

0.9763 atm * V = 0.0834 mol* 0.08206 atm.L/mol.K * 296 K

V = 2.07 L

Answer: 2.07 L

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