Question #f572e

Always start by writing a balanced chemical equation for the reaction that you have to work with.

In your case, you're dealing with a single replacement reaction in which zinc, the more reactive metal, displaces copper to form zinc sulfate and copper metal

#Zn_((s)) + CuSO_(4(aq)) -> ZnSO_(4(aq)) + Cu_((s))#

Notice that you have #1:1# mole ratios between all the species that take part in the reaction. In other words, regardless of how many moles of copper you have, you'll need an equal number of moles of copper sulfate and you'll produce equal numbers of moles of zinc sulfate and copper metal.

So, determine how many moles of zinc you have by using its molar mass

#2cancel("g") * "1 mole Zn"/(65.39cancel("g")) = "0.0306 moles Zn"#

Since copper sulfate is in excess, zinc will be your limiting reagent, i.e. it will be completely consumed by the reaction.

If 0.0306 moles of zinc are consumed, and you have a #1:1# mole ratio between zinc and copper, you'll automatically produce 0.0306 moles of copper.

#0.0306cancel("moles Zn") * "1 mole Cu"/(1cancel("mole Zn")) = "0.0306 moles Cu"#

Now use copper's molar mass to determine how many grams you'd produce

#0.0306cancel("g") * "65.55 g"/(1cancel("mole Cu")) = "2.006 g Cu"#

Since you only gave one sig fig for the 2-g mass of zinc, the answer will be

#m_"copper" = color(green)("2 g")#

Check out these links, there are many more solved problems you could use as examples

http://socratic.org/chemistry/stoichiometry/limiting-reagent

http://socratic.org/chemistry/stoichiometry/percent-yield

http://socratic.org/chemistry/stoichiometry/equation-stoichiometry

http://socratic.org/chemistry/stoichiometry/mole-ratios

Here are the basic steps to solve a theoretical yield problem.

1. Write the balanced equation for the reaction between #"Zn"# and #"CuSO"_4#.
2. Use the molar mass of #"Zn"# to convert grams of #"Zn"# to moles of #"Zn"#.
3. Use the molar ratio from the balanced equation to convert moles of #"Zn"# to moles of #"Cu"#.
4. Use the molar mass of #"Cu"# to convert moles of #"Cu"# to grams of #"Cu"#.

In summary, the conversions are

#"g Zn" stackrel(color(blue)("molar mass"))(→) "mol Zn" stackrel(color(blue)("molar ratio"))(→) "mol Cu" stackrel(color(blue)("molar mass"))(→) "g Cu"#

To 1 significant figure, you should get

#"2 g Zn" → "0.03 mol Zn" → "0.03 mol Cu" → "2 g Cu"#