HKCEE chemistry (help)

[匿名]

2007-04-20 12:31 am

1. 2NaN3(s)---->2Na(s)+3N2(g) what is the mass of sodium azide required to produce 72dm*3 of nitrogen at room temperature and pressure =24dm*3 (relative atomic masses :N=14.0,Na=23.0,molar volume of gas at room temperature and pressure =24 dm*3

2.which of the following gases contains the greastest number of molecules ?
a.50.0g of neon
b.50.0g of oxygen
3.50.0g of hydrogen chlorine
d.50.0g odf carbon monoxide

回答 (1)

魏王將張遼

2007-04-20 10:55 am

✔ 最佳答案

(1) 2 moles of sodium azide decompose to give 3 moles of nitrogen gas.
Therefore, according to the requirement in the question, 72/24 = 3 moles of nitrogen are needed to be produced.
Hence, 2 moles of sodium azide are needed and the mass is:
2 × (23 + 3 × 14) = 130 g
(2) First of all, we have to obtain the molar mass of the listed gases as follows:

Neon: Monatomic molecules, molar mass = 20 g/mol
Oxygen: Diatomic molecules, molar mass = 2 × 16 = 32 g/mol
Hydrogen chloride: HCl molecules, molar mass = 1 + 35.5 = 36.5 g/mol
Carbon monoxide: CO, molar mass = 12 + 16 = 28 g/mol
Therefore, in 50 g, neon will have the greatest number of moles and hence the number of molecules contained in this amount of gas will also be the greatest since 1 mole of gas contains 6.02 × 1023 molecules, regardless of monatomic or diatomic.

參考: My chemical knowledge

👍 0 👎 0