The ratios which are needed to determine the mass of oxygen produced from the decomposition of 10 grams of potassium chlorate are;
- 31.998 g O2 : 1 mole O2
- 3 mole O2 : 2 mole KClO3
- 112.55 g KClO31 mole KClO3
From stoichiometry;
- We can conclude that according to the reaction;
3 moles of oxygen requires 2 moles of KClO3 to be produced.
And from molar mass analysis;
- 31.998 g O2 is equivalent to 1 mole O2
- O2112.55 g KClO3 is equivalent to 1 mole KClO3
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Answer:
a) 381.2 g
b) 39916 g
c) 0.0013 lb mol
d) 29.6 g mol
Explanation:
The molecular weight (mw) of a compound is the mass of it per mole, so it's the ratio of the mass (m) per mole (n).
a) The molecular weight of one mol is found at the periodic table. So, for Mg, mw = 24.3 g/mol, for Cl = 35.5 g/mol, so for MgCl2, mw = 24.3 + 2*35.5 = 95.3 g/mol. The g mol is the mass divided by the molecular weight:
g mol = m/mw
4 = m/95.3
m = 381.2 g
b) The pound (lb) is a unity of mass, and the lb mol is a unity of the mass divided by the molecular weight. So, by the periodic table, the molecular weight of C3H8 is 3*12 (of C) + 8*1 (of H) = 44 lb/mol.
lb mol = m/mw
2 = m/44
m = 88 lb
1 lb = 453.592 g
So, m = 88*453.592 = 39916 g
c) The molecular weight of N2 is 2*14 (of N) = 28 lb/mol.
m = 16/453.592 = 0.0353 lb
lb mol = m/mw
lb mol = 0.0353/28
lb mol = 0.0013 lb mol
d) The molecular weight is 2*12 (of C) + 6*1(of H) + 1*16(of O) = 46 g/mol
3 lb = 1360.78 g
g mol = m/mw
g mol = 1360.78/46
g mol = 29.6 g mol
Answer: 873 kJ of energy will be required to break two moles of hydrogen gas.
Above chemical equation shows that two moles of hydrogen gas is reacting with one mole of oxygen gas to give two mole of water molecule.
Bond energy of H-H = 436kJ
According to reaction we have two moles hydrogen,so energy required to break the
H-H bond in two moles of hydrogen gas will be:
= 872kJ
872kJ of energy will be required to break H-H bond in two moles of hydrogen gas.
