[positiveion]

"What mass of hydrogen will be obtained if 100 cm^3 of 200 dm^-3 HCL are added to 4.86g of magnesium?"

Mg + 2HCl --> MgCl2 + H2

100cm^3 = 100 ml

100ml of 2.00 mol dm^-3 = 100 ml of 2.00mol/litre

100*2 = 200 moles

Therefore:
48g of Mg
7292g of Hcl

Mole calculations:
1 mol of Mg: 24.31
1 mol of Hcl: 36.46


48g of Mg:
48/24.31 = approx 2

Therefore the equation should be:
2Mg + 4Hcl --> 2 MgCl2 + 2H2

2H2 = 4 hydrogens

Hydrogen mol = 1.01

1.01*4 = 4.04

So I thought that the answer would be 4.04 g

HOWEVER - the answer is 2!

I don't understand it at all!

[Borek]

100 cm^3 of 200 dm^-3 HCL

Something wrong here. 100 cm³ of 0.200 M perhaps?

[positiveion]

Sorry it was 2.00

I just wrote typed that part wrong - it did not manifest itself in the calculations though

[Borek]

This is a limiting reagent problem.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents

[DrCMS]

100ml of 2.00 mol dm^-3 = 100 ml of 2.00mol/litre

100*2 = 200 moles

No NO NO! - if there are 2 moles in 1L how many moles in 100ml?

Therefore:
48g of Mg
7292g of Hcl

Mole calculations:
1 mol of Mg: 24.31
1 mol of Hcl: 36.46


48g of Mg:
48/24.31 = approx 2

No idea what you are doing here. 

You need to correctly calculate the no. of moles of HCl in 100ml 0f 0.2M HCl and the no. of moles of Mg in 4.86gMg.

Then using the 1st reaction equation you posted work out with reagent is limiting.

Now using that ammount calculate the ammount of Hydrogen formed.