一題reacting mass.............(empirical formula)
2008-07-26 7:02 am
13.7g of an oxide of metal X (relative atomic mass = 207 ) reacts with excess hydrogen to produce 1.44g of water. What is the empirical formula of the oxide?
回答 (2)
2008-07-26 7:31 am
✔ 最佳答案
Assume the empirical formula of the oxide isXnO
According to the chemical reaction between oxide and the H2
the mass of oxygen is
13.7 x [ 16/ (207n 16) ] = 1.44 x (16 / 18)
then
n = 3/4
Hence,
the empirical formula of the oxide is X3O4
參考: me
2008-07-26 7:22 pm
My answer will be:
The empirical formula of the metal oxide is X2On
The Chemical equation of this reactions will be:
X2On + nH2 -----> n H2O + 2X
(limiting reagent) (Excessive reagent) (Limiting reagent)
Since no. of mole water produced in the reaction = no. of mole of metal oxide
1/n x (1.44/18) = 13.7/ ((2*207) + 16n)
n = 2.667 (i.e. 3)
Finally, the empirical formula of the metal oxide X is X2O3
The empirical formula of the metal oxide is X2On
The Chemical equation of this reactions will be:
X2On + nH2 -----> n H2O + 2X
(limiting reagent) (Excessive reagent) (Limiting reagent)
Since no. of mole water produced in the reaction = no. of mole of metal oxide
1/n x (1.44/18) = 13.7/ ((2*207) + 16n)
n = 2.667 (i.e. 3)
Finally, the empirical formula of the metal oxide X is X2O3
參考: COMMON QUESTION APPEARED IN HKCEE CHEM PAST PAPER
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