Consider the reastion between NiS 2 and O 2 the following equation. Given 11.2g

Question

Consider the reastion between NiS2 and O2 the following equation. Given 11.2g of NiS2 will react with 5.43g of O2 to produce 4.86 g of NiO, what is the limiting reactant, the theoritical yield of nickel oxide, and the percent yield of nickel oxide? (HInt: consider the very 1st thing you should do when working with a chemical equation!)

2 NiS2 (s) + 5 O2 (g) --> NiO (s) + 4 SO2 (g)

1. Show your work for determination of limiting reactant.

2. Show your work for the theoretical yield of NiO

3. Show your work for percent yield of NiO

Explanation / Answer

first write the balanced equation

2 NiS2 (s) + 5 O2 (g) -----> 2NiO (s) + 4 SO2 (g)

from this reaction it is clear that 2 mol of NiS2 required 5 mol of O2

1 mol of NiS2 required 5/2 mol of O2

lets calculate the no of moles of each reactent = weight / molar mass

moles of NiS2

= 11.2 g / 122.8 g/mol = 0.0912 mol

no of moles of O2 = 5.43 g / 32 g/mol = 0.1697 mol

0.0912 moles of NiS2 required 5/2 (0.1697 mol ) = 0.4242 mol of O2 required

but actually we have 0.1697 mol

so limiting agent is O2

so we have to calulate the yield with respect to limiting agent

from the balanced equation

1 mol of O2 is giving 2/5 mol of NiO

according ly

0.1697 mol of O2 will give 2/5 (0.1697 mol) = 0.06788 mol of NiO

so moles of NiO formed theritically = 0.06788 mol

weight of NiO theriticall = moles x molar mass = 0.06788 mol x 74.7 g/mol = 5.0706 g

but actually they got is 4.86 g

now % of yield = (actual yield / theritical yield) x 100

= (4.86 g / 5.0706 g) x 100

= 95.84 %

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