J.R. S. answered 07/04/18
Tutor
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(145)
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
Let the mass of chalk in the 5 g sample = y grams
Mass of clay in the sample = 5 g - y g = 5-y grams
Since 1 mole of CaCO3 produces 1 mole of CO2 and the molar mass of CO2 = 44 g/mole and molar mass CaCO3 = 100 g/mole, one can relate mass of CO2 lost to the mass of chalk in the sample. Thus,.....
Mass CO2 lost = 44g/100g x y = 0.44y
Mass of water lost = (5-y)(0.11) = 0.55y
Total mass lost (as given in the question) = 1.1 g
Thus, we have 1.1 g = (0.55-0.11y) + 0.44y
0.33y = 1.1-0.55
y = 1.67 grams of chalk in the original sample
Percent chalk in original sample = 1.67 g/5 g (x100%) = 33.4%