I need help with stoichiometry?

Question

That is a sample question. If someone could answer it and guide me through it I'd be very thankful....

A 1.1 g sample of magnesium is treated with 9.5 g of sulfuric acid.

Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g)

(a) How many grams of hydrogen can be produced?

(b) If 0.060 g of hydrogen is actually obtained, what is the percent yield?

(c) Calculate the number of moles of excess reagent remaining at the end of the reaction.

Answer 1

a) work out the molecular weight for the compounds in question.
Then work out the number of moles in 1.1 g of Mg and 9.5 gg
H2SO4 then work out the limiting limit is the number of moles >
or < than the number of moles of Mg. Take the one with the
lowest limit and this is how many number of moles of H2 that
should be produced as is a ratio of (1:1). Then take the number
of moles and times the molecular weight of H2.

b) From a) Yield = (0.06 g / of theoric the weight of H2 produced)
x 100%

c) Number of moles of H2 made = 0.06 g /the molecular weight of
H2.

from a) Number of moles of Mg left = staring number of moles
of Mg - the number of moles of H2 produced.

from a) Number of moles of H2SO4 left = staring number of
moles of H2SO4 - the number of moles of H2 produced.

Answer 2

The reaction is already balanced, so just look at the molar quantities.

I don't have my periodic table handy, but this is the method.

1.1 gr of Mg / molecular mass of Mg = moles of Mg.

each mole of Mg gives the same # moles of H2 gas. Multiply by 2 grams per mole to get grams of H2. Divide 0.060 actual grams of H2 by the stoich.number created above to get the percent yield.