Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen , .
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon , .
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
The magnesium dissolves to form magnesium chloride and hydrogen gas is released.
Mg + 2HCl -----> MgCl2 + H2(g).
Answer:
2 KClO3 (s) = 2 KCl (s) + 3 O2 (g)
2.5 g x g
Explanation:
x g O2 = 2.5 g KClO3 x (1 mol KClO3) x (3 mol O2) x (32 g O2) = 0.98 g O2
(122.5 g KClO3) (2 mol KClO3) (1 mol O2)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
2.5 g x g
x g KCl = 2.5 g KClO3 x (1 mol KClO3) x (2 mol KClO3) x (74.5 g KCl) = 1.52 g KCl
(122.5 g KClO3) (2 mol KClO3) (1 mol KCl)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
x mol 10 mol
x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3
(3 mol O2)
Answer:
0.158 moles KMnO4
Explanation:
According to the Periodic Table,
K = 39.10 g/mol
Mn = 54.94 g/mol
O = 16.00 g/mol
KMnO4 = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol) = 158.04 g/mol
25.0 grams KMnO4 1 mole
----------------------------- x -------------------------- = 0.158 moles KMnO4
158.04 grams