A sample of gas occupies 300 ml at STP. Under what pressure would this sample oc
ID: 911195 • Letter: A
Question
A sample of gas occupies 300 ml at STP. Under what pressure would this sample occupy a volume of 150 ml at a temperature of 546 C?
Calculate the pressure exerted by 60.0 g C2H6 (ethane) in a 30.0-L vessel at 27 C.
When nickel is heated in carbon monoxide, a gaseous product is formed that has a density of 6.57 g/L at 750 torr and 40 C. The compound is analyzed to show 34.30% Ni, 28.1% C and 37.53% O. What is the molecular formula of the compound?
If 2.0 liters of N2 at 2.0 atm and 2.0 liters of H2 at 1.0 atm are put in a 4.0-liter vessel, what is the final pressure, temperature being constant?
Two gases, Ar and He, are contained in vessel. The mole fraction of He is 0.35 with a partial pressure of 1.15 atm. What is the partial pressure of Ar?
Given the reaction below, how many grams of iron are needed to produce 100 liter of hydrogen gas at STP?
3Fe(s) + 4H2O(l) Fe3O4 (s) +4H2 (g)
What volume of oxygen at 18 C and 750 torr can be obtained from the thermal decomposition of 100-g KClO3?
2KClO3 (s) 2KCl(s) +3O2 (g)
A student collects 2.63 L of O2 over water at 29 C and 769 torr by thermal decomposition of potassium chlorate. a) How many grams of oxygen are in the sample? b) How many grams of KClO3 decomposed to produce the oxygen?
2KClO3 (s) 2KCl (s)+ 3O2 (g)
Explanation / Answer
Answer – 1) We are given, volume, V1 = 300 mL, P1 = 1.00 atm, T1 = 273 K
V2 = 150 mL , T2 = 546 +273 = 819 K
We know, combine gas law
P1V1/T1 = P2V2/T2
So, P2 = P1V1T2 / T1V2
= 1.00 atm * 300 mL * 819 K / 150 mL * 273 K
= 6.00 atm
6.00 atm pressure would this sample occupy a volume of 150 ml at a temperature of 546 C
2) Given, mass of ethane , C2H6 = 60.0 g , T = 273 +273 = 300 K , V = 30.0 L
First we need to convert the mass to moles
Moles of C2H6 = 60.0 g / 30.07 g.mol-1
= 2.0 moles
We know Ideal gas law
PV = nRT
So, P = nRT/V
= 2.00 moles * 0.0821 L.atm.mol-1.K-1 *300 K / 30.0 L
= 1.64 atm
So, 1.64 atm pressure exerted by 60.0 g C2H6 (ethane) in a 30.0-L vessel at 27 C.
3) We are given, density of gas = 6.50 g/L , P = 750 torr /760 = 0.989 atm
T = 40 +273 = 313 K
So molar mass of gas
Density = PM/RT
So, M = density * RT / P
= 6.50 g.L-1* 0.0821 L.atm-1*313 K / 0.989 atm
= 169.3 g/mol
Assume, 100 g of sample
So, Ni = 34.30 g , C = 28.1 g , O = 37.53 g
Moles of Ni = 34.30 g / 58.693 g.mol-1
= 0.584 moles
Moles of C = 28.1 g / 12.01 g.mol-1
= 2.34 g
Moles of O = 37.53 g / 15.998 g.mol-1
= 2.34 g
So, moles of Ni is lowest
Ni = 0.584 / 0.584 = 1
C = 2.34 / 0.584 = 4.0
O = 2.34 / 0.584 = 4.0
So empirical formula is NiC4O4
Molecular formula = n * empirical formula
n = molecular mass / empirical formula mass
= 169.3 / 170
= 1
So molecular formula = 1 * NiC4O4
= NiC4O4
4) We are given, V1 = 2.0 liters of N2 , P1 = 2.0 atm, V2 = 4.0 L, P2 = ?
V1 =2.0 liters of H2 , P1 = 1.0 atm , P2 = 4.0-liter
Partial pressure of N2
P2 = P1V1/V2
= 2.0 L * 2.0 atm / 4.0L
= 1.0 atm
Partial pressure of H2
P2 = P1V1/V2
= 2.0 L * 1.0 atm / 4.0L
= 0.5 atm
So total pressure = 1.0 + 0.5 = 1.5 atm