can you help me with this chem problem?

Question

A sample of calcium metal with a mass of 2.00 was reacted with excess oxygen the following equation represent the reaction 2Ca+O2-->2CaO
the isolated product weighed 2.26g what was the percentage yield of the reaction

Answer 1

atomic mass of Ca/(atomic mass of Ca + atomic mass of O) =
% of Ca.

2.26g * % of Ca/2.0g = solution
===============================
40/56 = 5/7
5/7 * 2.26 g / 2.0g = 80.7%

Answer 2

atomic mass of Ca = 40g
no. of moles in 2g of Ca = 2/40
= 0.05
since 2 moles of Ca react to produce 2 moles of Ca0
the ratio is 1:1
so no. of moles of Ca0 produced should rightfully be 0.05 moles.

molecular mass of Ca0 = 40 + 16
= 56g
weigh of 0.05 moles of Ca0 = 56 * 0.05
= 2.8g

2.8g is the theoretical mass but only 2.26 g was produced
so percentage yield
= 2.26 / 2.8 * 100%
= 80.7142....
= 80.7% (3 sig. fig.)

hope that helps! :)

Answer 3

You know how much product you got (2.26g). Find how much product you COULD have gotten:

2 g Ca | 1 mole / 40.1 g | = 0.05 moles Ca

Your equation says that for every 2 moles of Ca reactant used, you could get 2 moles of CaO product. So for 0.05 moles of Ca, you could get 0.05 moles of CaO.

Add the atomic masses of Ca and O to find the molecular mass of CaO:
40.08 + 16.0 = 56.08 g per mole of CaO

Now find the mass of 0.05 moles of CaO:

0.05 moles | 56.08 g/mole | = 2.804 g CaO

The percent yield is the actual yield (2.26 g), divided by the theoretical yield (2.804 g), times 100:

% yield = (2.26 / 2.804) x 100% = 80.6%

Answer 4

So, you have 2,00g of Ca:
2,00g Ca*(1mol Ca/40,08g Ca)= 0,0500 moles of Ca
2 moles of Ca react with 1 mol of O2 to give 2 moles of CaO, so:
0,0500moles Ca*(2 moles CaO/2 moles Ca)(56,08g CaO/1mol CaO)= 2,80 g of CaO (that is the amount of CaO you are expected to get)
(2,26g CaO/2,80g CaO)*100= 80,7% yield.