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Chem Form4 masses and volume計算
2009-08-14 11:53 pm
1.Calculate the volume of carbon monoxide(measured in r.t.p.) needed toreduce 111.0g of litharge (PbO) to lead .Also calculate the volume of the gaseous product (at r.t.p.).
ps:(我有一本chem mc,叫A New Certificate Chemistry Multiple Choice Questions ,作者是 C.Y.Tong 的.我冇mc ans,如果你們知道有,可否告訴我網址，ｏｒ你們有，可否打上黎給我？謝謝！～
ps:(我有一本chem mc,叫A New Certificate Chemistry Multiple Choice Questions ,作者是 C.Y.Tong 的.我冇mc ans,如果你們知道有,可否告訴我網址，ｏｒ你們有，可否打上黎給我？謝謝！～
回答 (2)
2009-08-15 5:07 am
✔ 最佳答案
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
參考: Physics king
2009-08-20 11:12 pm
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!~
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!~
參考: Me
收錄日期: 2021-04-19 15:11:21
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