When 15 grams of copper (ii) chloride (cucl2) reacts with 20 grams of sodium nitrate (nano3), 11.3 grams of sodium chloride (nacl) is obtained. what is the percent yield of this reaction? cucl2 + 2nano3 cu(no3)2 + 2nacl
options are
a) 11.3%
b) 13.1%
c) 26.2%
d) 43.45%
e) 86.9%

86.9%  

Explanation:

Moles of CuCl₂ = mass of CuCl₂/molar mass of CuCl₂= 15g / 135.45 g/ml = 0.11 moles

Moles of NaNO₃ = mass of NaNO₃/molar mass of NaNO₃ = 20 g / 84.9947 g/ml = 0.2353 moles

CuCl₂ is the limiting agent as it has lowest number of moles.  

CuCl₂ will dictate the yield of NaCl.  

From the balanced equation,  

1 mole of CuCl₂ = 2 moles of NaCl

0.11 moles of CuCl₂ = 0.22 moles of NaCl

Mass of NaCl = moles of NaCl x molar mass of NaCl  = 0.22 mol x 58.44 g/mol = 12.8568 g

Percent yield = actual yield/ theoretical yield x 100  

Percent yield is the ratio of the actual yield to the theoretical yield  

Given,  

Actual yield NaCl = 11.3 g  

Theoretical yield NaCl = 12.8568 g   ~ 13 g

Percent yield =  11.3 g / 13 g   = 86.9%  

these nuts

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"