How many grams of NO_2, nitrogen dioxide, are contained in 0.500 L of the gas at

Question

How many grams of NO_2, nitrogen dioxide, are contained in 0.500 L of the gas at STP? Calculate the density of NO_2 gas, in grams per liter, at 110 degree C and 12 atm. Calculate the molar mass of a gas if 0.125 g of the gas occupies 93.3 mL at STP. On a spring morning (15 degree C) you fill your tires to a pressure of 2.25 atmospheres. As you ride along, the tire heats up to 45 degree C from the friction on the road. What is the pressure in your tires now? What volume of H_2 is formed at STP when 6.0 g of Al is treated with excess NaOH? 2NaOH + 2Al + 6H_2 O rightarrow 2NaAl(OH)_4 + 3H_2 (g)

Explanation / Answer

a) at STP 273 K temperature and 1 atm pressure

number of mol of NO2 (n) = PV / RT ............. { R = 0.082 L atm/mol K}

n = (1 atm x 0.50 L / 0.082 x 273 K) = 0.02234 moles

number of gram of NO2 = ( moles x molar mass ) = ( 0.02234 x 46 g/mol ) = 1.027 g

b) Density = (Pressure)(Molar Mass) / (R)(Temperature in K )

density = (12 atm x 46 g/mol) / (0.082 x 383 K) = 17.57 g/L

c) mol = mass / molar mass

no. of mol of gas = (1 atm x 0.0933 L) / (0.082 x 273 K) = 0.00417 mol

molar mass = (0.125 g / 0.00417 mol) = 30 g/mol

d) volume are constant in tire

P1 / T1 = P2 / T2

P2 = (P1 / T1) T2

P2 = (2.25 atm / 288 K ) x 318 K

P2 = 2.48 atm

e) 2 mol Al react with 6 mol H2O produce 3 mol H2 gas

so no. of mol of Al in 6.0 g = (6.0g / 27 g/mol ) = 0.22 mol

0.22 mol Al react with 0.66 mol H2O produce 0.33 mol H2

volume of H2 ( V ) = nRT / P

V = (0.33 mol x 0.082 x 273 K / 1 atm) = 7.389 L

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.