ks People Window Help 100% . Sur General Chemistry /Mms/quiz ral Chemistry 1-CH.

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ks People Window Help 100% . Sur General Chemistry /Mms/quiz ral Chemistry 1-CH. saling Learning I Login Page × Conversions-Chem pendix x" wwwmacmilanhigheredeon ttemptiquizstartframe.danou-86984&isprv;_&drc;=0&q;ì-2869 2 1 &cfgl-0;&dnbao; 1 1 Presley Folkerts LEGE eCampus Course l Content l Dropbox 1 Progress l Resources 1 Tools Exp 3 Prelab Gravimetric Determination of the Water of Hydration of Epsom Salt*-Quilz me Left: 1:48:16 Presley Folkerts: Attempt 1 Save Question 2 (2 points) You are given a crushed sample that is a mixture of Limestone (calcium carbonate), lime (calcium axide), and sand. The calcium carbonate, or limestone, is the only material present in the mixture that will decompose when heated. You subject a 6.3255 g sample of the mixture to strong heating and after the sample reaches constant mass (no more mass is lost with additional heatingl, the sample has a final weight of 3.9747 g. What is the percentage of calcium carbonate present in the original mixture? MW of calcium carbonate -100.1 g/mol) Equation for reaction Cacos)Cao) cog Your Answer Answer units Save Question 3 (2 points) of

Explanation / Answer

Ans. Step 1: Calculate Moles of CO2 evolved:

Upon heating CaCO3, all the mass lost is in form of gaseous CO2.

So, loss in mass must be equal to the mass of CO2 evolved.

Now,

            Mass of CO2 evolved = Initial sample mass – Mass of dried sample

                                                = 6.3255 g – 3.9747 g

                                                = 2.3508 g

            Moles of CO2 evolved = Mass / Molar mass

= 2.3508 g / (44.01 g/ mol)

= 0.053415 mol

# Step 2: Calculate CaCO3 following stoichiometry:

Balanced reaction:    CaCO3(s) ---heat----> CaO(s) + CO2(g)

According to the stoichiometry of balanced reaction, 1 mol CaCO2 yields 1 mol CO2.

Therefore, moles of CaCO3 in sample must be equal to the moles of CO2 evolved.

So,

            Moles of CaCO3 = 0.053415 mol = Moles of CO2 evolved

Now,

            Mass of CaCO3 in sample = Moles x Molar mass

                                    = 0.053415 mol x (100.1 g/ mol)

                                    = 5.3468 g

# % CaCO­3 in sample = (Mass of CaCO3 / Initial mass of sample) x 100

                                    = (5.3468 g/ 6.3255 g) x 100

                                    = 84.53 %

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