Given the equation 2KCIO3(s)=2KCI(s) + 3O2(g). A 3.00-g sample of KCIO3 is decomposed and the oxygen at 24 degrees C and 0.982 a
1 answer:
You might be interested in
Answer: The empirical formula is .
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of = 12.24 g
Mass of = 2.505 g
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 12.24 g of carbon dioxide, = of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 2.505 g of water, = of hydrogen will be contained.
Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H=
Moles of O=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =
For O =
The ratio of C : H : O = 3: 3: 1
Hence the empirical formula is .