Thirty points to who ever can answer this problem

during an experiment, 104 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. if the percent yield of the reaction was 80.15%, what was the actual amount of calcium chloride formed?
caco3 + hcl → cacl2 + co2 + h2o

a. 90.1 grams
b. 92.4 grams
c. 109.2 grams
d. 115.3 grams

B. 92.4 g

Explanation:

the balanced equation for the reaction is as follows

CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O

molar ratio of CaCO₃ to CaCl₂ is 1:1

number of CaCO₃ moles reacted - 104 g / 100 g/mol = 1.04 mol

therefore number of CaCl₂ moles reacted - 1.04 mol

mass of CaCl₂ expected to be formed = 1.04 mol x 111 g/mol = 115.44 g

percentage yield = actual yield / theoretical yield x 100 %

theoretical yield = 115.44 g

percentage yield = 80.15 %

substituting these values in the above equation

80.15 % = actual yield / 115.44 g x 100 %

actual yield = 92.5 g

therefore answer is B. 92.4 g

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.