How many grams of nickel must react with an excess volume of sulfuric acid to yield 0.500 g of nickel(ii) sulfate?

Ok so put firstly the reaction in a chemical form ->

Nickel chemical symbol is Ni
Sulfuric acid chemical formula is H2SO4
Nickel (II) sulfate's chemical formula is NiSO4

Nickel + Sulfuric Acid -> Nickel (II) Sulfate

Ni + H2SO4 -> NiSO4 + H2

Ok now check if the equation is balanced.

1 Ni , 2 H2 ,1 S , 4 O on the RHS ( Right hand side )

1 Ni , 2 H2 , 1 S , 4 O on the LHS ( Left hand side )

Therefore it is balanced.

Ok so we want to find how many grams of Ni ( Nickel) will give us .5 g of NiSO4.
In order to find out the answer we need to convert the information we know into moles as the reactions are always in ratio ( where you can only work with ratio when you have moles)

Sulfuric acid is in excess therefore we don't take it into account We only look at the ratio between Nickel and Nickel (II) Sulfate ( also we don''t need H2 ( Hydrogen gas) therefore we don't work it out ( but it is possible to do so) )

Ok lets convert 0.5 grams of NiSO4  into moles.

The following conversion will make it easier for you to convert.

Mass in grams divided by R.M.M of an element/molecule/compound ( Relative Molecular Mass ) -> Moles

Moles multiplied by R.M.M -> Grams

Lets fine out the R.M.M if NiSO4 
Ni-59 ( Rounded)
S-32
4 of O - 16 * 4 = 64

                             155
Ok now divide this into the mass.
0.5 / 155 = 0.0032 moles ( rounded )

Ni + H2SO4 -> NiSO4 + H2
x moles  excess  0.0032moles

Ok the ratio between Ni and NiSO4 is 1:1 
Therefore 
x:0.0032
x=0.0032

Now we know that Ni has 0.0032 moles to find the mass multiply by R.M.M of Ni.

0.0032 * 59 = 0.19 grams of Nicel must react ( rounded).

Hope this helps :).