Anonymous

# 中四化學計算

chemical equation:2Al(s)+Fe2O3(s)---->Al 2O3(s)+2Fe(s)

Calculate the mass of aluminium required to react compeletely with 40.0g of iron(III)oxide.

(Relative atomic mass: O=16.0 ,Al=27.0 ,Fe=56.0)

217g of manganese(IV)oxide are treated with hydrochloric acid containing 274g of hydrogen chloride. Determine

1)the limiting agent

2)the mass of chlorine producted;and

3)the mass of excess reagent left after the reaction

(Relative atomic masses; H=1.0, O=16.0 ,Cl=35.5 ,Mn=55.0)

miss不斷用英文講,我知我要用英文,不過我希望我可以明左點解先

Rating

我幫下你啦.

記住, 做呢種calculation一定要將mass (kg, g)轉晒moles先, 唔係點計都唔會岩!

1. 問題問幾重既aluminium會同40.0g既iron(III)oxide完全反應:

首先, 揾出40.0g既iron(III)oxide等於幾多moles:

No. of moles of iron(III) oxide = 40.0 / (56x2+16x3) = 0.25

然後看化學式:

2Al(s)+Fe2O3(s)----&gt;Al 2O3(s)+2Fe(s)

1mole既iron(III) oxide 會同2moles既aluminium反應, 所以如果你有0.25moles既iron(III) oxide就會同0.50moles既aluminium反應!

再看0.50moles既aluminium有幾重:

Mass of alumnium = 0.50 x 27.0 = 13.5g

答案就係13.5g!

2. 原理差唔多, 但係呢次你要揾邊種反應晒同邊種用唔晒, 反應晒果種就是limiting reagent:

先揾217g of manganese(IV)oxide有幾多moles:

No of moles of manganese(IV)oxide = 217/(55+16x2) = 2.494

再揾274g of HCl有幾多moles:

No of moles of HCl = 274/(1+35.5) = 7.507

然後看化學式:

MnO2+4HCl(aq)---&gt;MnCl2(aq)+Cl2(g)+2H2O(l)

1mole既MnO2會同4moles既HCl反應, 你有2.494moles既MnO2同埋7.507moles既HCl, 明顯MnO2多左出黎! 因為7.507moles既HCl淨係會同(7.507/4 = 1.877)mole既MnO2反應, 但係你有2.494moles既MnO2! 所以, HCl係你既limiting reagent!

先答(iii):

我地多左(2.494 - 1.877 = 0.617)mole既MnO2出黎, 所以

Mass of excess reagent = 0.617 x (55+16x2) = 53.7g

再答(ii):

我地用limiting reagent黎計, 因為佢反應晒, 我地唔想果D多左出黎既野搞亂我地:

化學式話4moles既HCl可以搞到1mole既chlorine gas出黎, 所以:

No of moles of Cl2 produced = 7.507 / 4 = 1.877

轉番做mass就搞掂啦!

Mass of Cl2 produced = 1.877 x (35.5x2) = 133.3g