chem's question

2008-04-13 4:06 am

1A few years ago, many alcoholics in a small village in China had drunk
some bogus alcohols manufactured by business culprits and many blinded
and even some died. Which of the following is the most possible substance
present in the bogus alcohols?

A.Methane
B.Methanol
C.Ethanol
D.Ethanoic acid

2The following reactions can be applied to produce oxygen in laboratory:

Reaction 1: 2KClO3  2KCl + 3O2 upon heating with catalyst

Reaction 2: 2HgO  2Hg + O2 upon heating

Reaction 3: 2Na2O2 + 2H2O  4NaOH + O2 upon heating

If the same total mass(es) of reactant(s) are used, the ascending order of
mass of oxygen produced would be:

A.Reaction 1 < Reaction 3 < Reaction 2
B.Reaction 1 < Reaction 2 < Reaction 3
C.Reaction 3 < Reaction 2 < Reaction 1
D.Reaction 3 < Reaction 1 < Reaction 2
3.100 cm3 of a gas X2 react completely with 300 cm3 of hydrogen to form 200
cm3 of a gaseous hydride. If all measurements are made at room temperature,
the formula of the gaseous hydride is

A.X3H2.
B.XH4.
C.X2H3.
D.XH3.

回答 (1)

Uncle Michael

2008-04-13 6:56 am

✔ 最佳答案

1. The answer is B.

Methanol is toxic, which can cause blindness or can even kill.

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2. There is no answer.
2. The answer should be: reaction 2 < reaction 3 < reaction 1

Molar mass of KClO3 = 39 35.5 + 16x3 = 122.5 g mol-1
Molar mass of HgO = 200.6 + 16 = 216.6 g mol-1
Molar mass of Na2O2 = 23x2 + 16x2 = 78 g mol-1
Molar mass of H2O = 1x2 + 16 = 18 g mol-1
Molar mass of O2 = 16x2 = 32 g mol-1

In reaction 1, 2 mol of KClO3 gives 3 mol of O2.
(122.5x2 = 245 g) of reactant gives (32x3 = 96 g) of O2.
1 g of reactant gives (96/245 = 0.392 g) of O2.

In reaction 2, 2 mol of HgO gives 1 mol of O2.
(216.6x2 = 433.2 g) of reactant gives 32 g of O2.
1 g of reactant gives (32/433.2 = 0.074 g) of O2.

In reaction 3, 2 mol of Na2O2 and 2 mol of H2O give 1 mol of O2.
(78x2 + 18x2 = 192 g) of reactants give 32 g of O2.
1 g of reactants give (32/192 = 0.167 g) of O2

==========
42.
The answer is D.

Let XnHm be the formula of the gaseous hydride.
nX2(g) mH2(g) → 2XnHm

For all gases at room condition, mole ratio = volume ratio (Avogadro’s law).
Mole ratio X2 : XnHm = n : 2 = 100 cm3 : 200 cm3
Hence, n = 1

Mole ratio H2 : XnHm = m : 2 = 300 cm2 : 200 cm3
Hence, m = 3

Answer: Formula = XnHm = XH3

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