Hãy tìm công thức hóa học của những oxit có thành phần khối lượng như sau S:50%; C:42,8%; Mn:49,6%; Pb:86,6%
1 answer:
Answer:
Find the chemical formulas of oxides that contain the following composition:
S:50%; C:42,8%; Mn:49,6%; Pb:86,6%
Explanation:
In the given oxide of sulphur,
50%S is present that means remaining 50% is O.
Divide the % with atomic mass:
S O
50/32= 1.5625 50/16=3.125
divide with smallest ratio:
1.5625/1.5625=1 3.125/1.5625=2
Hence, the formula of oxide is .
Similarly:
Carbon is 42.8% means the reamining % will be O that is 57.2%.
C O
42.8/12=3.575 57.2/16=3.575
Divide with smallest ratio:
3.575/3.575=1 3.575/3.575=1
Hence, the formula of oxide is CO.
Mn is 49.6% means the remining is % is O that is 50.4%.
First divide with atmic mass of respective element, then get the smallest ratio.
That gives the mole ratio of each constituent atom.
Mn O
49.6/55.0=0.90 50.4/16=3.15
0.90/0.90=1 3.15/0.90=3.5
Multiply with two
2 7
Hence, the formula becomes:
Pb is 86.6% means ---remaining 13.4% is O.
Pb O
86.6/207 = 0.418 mol 13.4/16=0.85
0.418/0.418=1 0.85/0.418=2.0
Hence, the formula is:
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Answer:
a) No molecules of hydrogen
b) four molecules of ammonia
c) four left molecules of nitrogen.
Explanation:
The balanced reaction between nitrogen and hydrogen molecules to give ammonia molecules is:
Thus one molecule of nitrogen will react with three molecules of hydrogen to give two molecules of ammonia.
We have six molecules of each nitrogen and hydrogen in the closed container and they undergo complete reaction it means the limiting reagent is hydrogen. For six molecules of nitrogen, eighteen molecules of hydrogen will be required.
So six molecules of hydrogen will react with two molecules of nitrogen to give four molecules of ammonia.
The product mixture will have
a) No molecules of hydrogen
b) four molecules of ammonia
c) four left molecules of nitrogen.