Chem Form4 masses and volume計算

Chemical equation:

CO(g) + PbO(s) → CO2(g) + Pb(s)

Molar mass of litharge

= 237.0 + 16.0

= 253.0 gmol-1

Number of mole of litharge

= Mass / Molar mass

= 111.0 / 253.0

= 0.439 mol

From the equation, molar ratio of litharge to carbon monoxide = 1 : 1

So, number of mole of carbon monoxide required = 0.439 mol

1 mole of gas occupies 24.0 dm3 of volume in r.t.p.

So, volume of carbon monoxide required

= 24.0 X 0.439

= 10.5 dm3

From the equation, molar ratio of carbon dioxide to carbon monoxide is 1 : 1

So, the volume of the gaseous product, CO2 = 10.5 dm3

2009-08-20 15:23:38 補充：
Molar mass of PbO = 207 + 16 = 223 gmol^-1
No. of mole of PbO = 111 / 223 = 0.498 mol
Volume of CO required = 24.0 X 0.498 = 11.9 dm^3
So, the volume of the gaseous product = 11.9 dm^3

CO(g) + PbO(s)-----> Pb(s) + CO2(g)

no. of mole of PbO=111/(207+16)
=11.9dm^3
because mole ratio of PbO : CO is 1:1
therefore volume of gaseous product is also 11.9dm^3.

2009-08-20 15:13:54 補充：
不是上面的10.5dm^3呀!~