[abdi]

Hi
I have a problem with questions about Calculating from Reacting masses
Here is one

During steel making, sulphur is removed from molten iron by adding magnesium.If a 100 tonne batch of molten iron contains 0.25% sulphur, how many kilograms of magnesium are needed to remove it?

I started of with doing the equation, Mg + S ------> MgS
I then calculated the molar mass, Mg=24 S=32,  24+32= 56

After this Im stuck

[Borek]

What is mass of the sulfur that have to react with Mg?

[abdi]

Its not mentioned

[Borek]

Yes it is. Read the question carefully.

[Alberto_Kravina]

Yup it is given, look deeper....

To solve your problem:

n(Mg):n(S)=1:1

n(Mg)=n(S)

If you have x moles of elementar sulfur you need exactly x moles of magnesium, so you should calculate how many moles of sulfur react, and after that you have to calculate how many kilograms of Magnesium correspond to the number of sulfur moles.

The whole problem is based on the equation n=m/M

[abdi]

do i calculate it like this for the sulfur

n=m/mr , 0.25/32

     

[Alberto_Kravina]

Nono!

The calculation to find out how much sulfur there is goes on like this:

100 Tonnes contain 0.25% sulfur => 100 tonnes are 100000 kg => 0.25% of this 100 tonnes are sulfur.

0.25% of 100000 kg are x kg

100% = 100000kg
0.25% = x kg => This is the mass of the sulfur, if you have this mass you can calculate how much magnesium you need.

Simple eh?

PS: the amount of sulfur is in percent! Not in kilograms or grams!

[Chrataxe]

Why would it be in percent?  The percentage is already given (25%).  So, multiply .25 by the mass and you will get a mass.

[Donaldson Tan]

what is the mass of sulphur? how many moles of sulphur is that?

the stoichiometric ratio is such that 1 mole of sulphur reacts with 1 mole of magnesium.

if there is 5moles of sulphur in the molten iron, then you will need 5moles of magneium to remove the sulphur from the molten iron.

next, you are required to find the equivalent mass (kg) of the number of moles of sulphur available in the molten iron.