The combustion reaction of propane would be expressed as:
C3H8 + 5O2 = 3CO2 + 4H2O
To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:
moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O
Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.
Answer:
572 g
Explanation:
Molar mass is the mass of 1 mol of an element or compound
molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound
molar masses of each element making up lithium sulphate
Li - 7 g/mol
S - 32 g/mol
O - 16 g/mol
molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )
molar mass = 110 g/mol
mass of 1 mol of Li₂SO₄ is 110 g
therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g
mass is 572 g
There are less sharks farther from coral reefs.
Answer:
The pressure in that cylinder = 1.12atm
Explanation:
We use general gas law to calculate it. General gas law is gotten by combining Boyle's law, Charles' law and Avogadro's law. Thus
P = nRT/V
Where n = number of moles
R = the gas constant
T is the Temperature, V is the volume and P is the pressure.
Given: T = 319K, V = 24L, R = 0.0821 L.atm/K.mol
The first step is to find n using
n = mass of O2/molar mass of O2
=32.7/32
=1.0219
Now, using P =nRT/V
P = 1.0219 ×0.0821×319÷24
Therefore P = 1.12atm
