Stoichiometry of gases 3 questions

Question

5. How many grams of water can be produced from the complete reaction of 3.44 L of oxygen gas, at STP, with hydrogen gas?

 

6. How many liters of gaseous carbon monoxide at 27°C and 0.247 atm can be produced from the burning of 65.5 g of carbon according to the following equation?
2C(s) + O2(g) → 2CO(g)

 

7. How many liters of H2 gas at STP can be produced by the reaction of 4.60 g of Na and excess water, according to the following equation?
2Na(s ) + 2H2O(l ) → H2(g) + 2NaOH(aq)

Answer 1

Hi Lindsey,

5.

Standard temperature and pressure (STP) is defined in Chemistry as

T = 273.15 K , P = 101325 Pa.

We know the volume of oxygen so we can calculate the number of moles of oxygen at the given lab condition.

V = 3.44 L (1m³/1000L) = 3.44x10^-3 m³

Treating oxygen as an ideal gas,

PV = nRT,  R = 8.3145 J/mol K

n = PV/RT

   =(101325 Pa)(3.44x10^-3m³)/(8.3145J/mol·K * 273.15 K)

   = 0.1535 mole

2H₂ + O₂ -----> 2H₂O

molecular mass of H₂O = 2*1.008+15.999 =18.015 g/mole

From this equation, 1 mole of Oxygen gas produces 2 moles of water,

number of moles of H₂O = (2 moles H₂O/1 mole O₂)(0.1535 mole O₂)

                                    = 0.307 mole

mass of H₂O produced = (molecular mass of H₂O)(moles of H₂O)

                                 = (18.015 g/mole)(0.307 mole)

                                 = 5.53 g

6.

mass of Carbon = 65.5 g, T = 27°C = 300.15 K,

P =0.247 atm = (0.247 atm)(101325 Pa/1 atm) = 25027.275 Pa

2C(s) + O₂(g) ---> 2CO(g)

2 moles of solid C  burned in oxygen produces 2 moles of CO gas (ratio= 1:1)

Number of moles of C = (65.5g)/(12.011 g/mole) = 5.453 moles 

Number of moles of CO is n = 5.4533 moles

PV = nRT

V = (nRT)/P

   = (5.4533 mole*8.3145 J/mol·K*300.15 K)/25027.275Pa

   = 0.5438 m³

   = (0.5438m³)(1000 L/ 1 m³)

   = 543.8 L

7.

2Na(s) + 2H₂O(l) --->2NaOH(aq) + H₂(g) 

2 moles of Na reacts with 1 moles H₂

The ratio of the number of moles of Na to H₂ = 2:1

Number of moles of Na = 4.60g/22.989g/mole =0.2000 mole

Number of moles H₂ = (1/2)(number of moles of Na

                              = 0.1000 mole

n = 0.1000 mole, P =101325 Pa, T = 273.15 K, V = ?, R = 8.3145 J/mol K

V = nRT/P = (0.1000 mole*8.3145 J/mol K*273.15 K)/101325 Pa

V = 0.00224 m³(1000 L/1 m³),            1 m³ = 1000L

Volume of hydrogen gas is 2.24 L