Stoichiometry Question?

Question

Mg + 2HCL --> MgCL2 + H2

What mass of hydrogen at STP is produced from the reaction of 50.0 g of Mg and the equivalent of 75 g of HCL?

How much of the excess reagent in problem 3 is left over?


Please show all of your work.

Answer 1

moles Mg = 50.0 g / 24.3 g/mol = moles Mg

moles HCl = 75 g / 36.5 g/mol = moles HCl

1 mole Mg reacts with 2 moles HCl (given in equation)

Answer 2

Make sure that you understand the procedure to do all stoichiometrical problems:
step 1:
calculate how many more of each reactant to see how much is available. don't get carried away by mass because sometimes the product with more mass can surprise you by being the one in shortage:
moles of Mg available=mass available/Mr
=50/24.3050
=2.0572 moles
moles of HCl available=mass of HCl/Mr
=75/(1+35.4527)
=2.0575moles

Now look at the equation:
it tell us that for each mole of Mg we require 2 moles of HCl in order to produce 1 mole of MgCl2 and another mole of H2 gas.
1 mole of Mg needs 2 moles of HCl
2 moles of Mg would need 4 moles of HCl
and
1 HCl mole of will need 1/2 mole of Mg and
2 HCl mole of will need 1 mole of Mg

since we have only 2 .0575 moles of HCl, we have less of it because we needed 4 to react with all Mg available
therefore HCl in this case is called a limiting reagent, and we will have some Mg left over as they won't have any HCl to react with.
we now seal with the reactant which is in shortage because when it gets used up, the reaction will no longer go on.

2 moles of HCl will produce 1 mole of H2 gas
therefore mass of H2 produced= moles*Mr
=1*1.007
=1.007 grams

moles of Mg remained= moles originally there- moles used
=2.0572-(2.0275/2)
=1.02845
therefore the mass of Mg remained= moles remained*Mr
=1.02845*24.30
=24.99 grams

hope this is helpful