N h2o = mh20/Mh20
= 8.0g/18g/mol
=0.44 mol
therefore n O = 0.44mol
therefore n H2 = 0.88 mol
m H2 = n H2 * M H2
= 0.88 mol * 2 g/mol
=1.76 g
Answer:
c. 8.1 L
Explanation:
Given that:-
Moles of oxygen gas = 0.50 mol
According to the reaction shown below as:-

3 moles of oxygen gas on reaction gives 2 moles of ozone
Also,
1 mole of oxygen gas on reaction gives 2/3 moles of ozone
So,
0.50 mole of oxygen gas on reaction gives
moles of ozone
Moles of ozone = 0.3333 mol
Pressure = 1 atm
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25.0 + 273.15) K = 298.15 K
Volume = ?
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V = 0.3333 mol × 0.0821 L.atm/K.mol × 298.15 K
⇒V = 8.1 L
Lar mass of Ca<span> = 40.08 </span>grams/mole 77.4 g Ca<span> * ( 1 </span>mole Ca<span>/ 40.08 ... n = m / M 1mol </span>Ca<span>weights 40 gmol-1 n = 77,4 / 40 = 1.93 </span>mol<span>.</span>
Answer:
P = 14.1 atm
Explanation:
Given data:
Mass of methane = 64 g
pressure exerted by water vapors = ?
Volume of engine = 24.0 L
Temperature = 515 K
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O + energy
Number of moles of methane:
Number of moles = mass / molar mass
number of moles = 64 g/ 16 g/mol
Number of moles = 4 mol
Now we will compare the moles of water vapors and methane.
CH₄ : H₂O
1 : 2
4 : 2/1×4 = 8 mol
Pressure of water vapors:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
P = 8 mol × 0.0821 atm.L/mol.K× 515 K / 24.0 L
P = 338.25 atm.L/ / 24.0 L
P = 14.1 atm