what is the percent by mass of CaCO3 in the 1.000 gram sample of dolomite?

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Dolomite is mixture of CaCO3 and MgCO3. When 1.000 g sample of this limestone is decomposed by heating, .268 L of Cao2 is produced measured @ 750. Torr and 20 C. Heating a metallic carbonate always produces carbon dioxide and a metallic oxide. It was determined that the original sample of the limestone contained 160.0 mg of calcium. What was the % by mass of CaCO3 in the original sample.

Sep 1st, 2015

268 L of CO2 is produced measured @ 750. Torr and 20 C.

n=pv/rt

62.364 L Torr K-1 mol-1

n= 750*268/62.364*293K

n = 11 MOLES

TOTAL WEIGHT OF CO2 = 44* 11 =484grams

caco3 heat CaO +Co2

PLEASE REVIEW QUESTION AND CONFIRM IF ALL DETAILS ARE CORRECT

Sep 1st, 2015

Your question appears ambigous . Are all details correct?

is this correct   Cao2 ? or should it be Co2

Sep 2nd, 2015

iGNORE PREVIOUS IT IS NOT IMPORTANT

160.0 mg of calcium. What was the % by mass of CaCO3 in the original sample.

IF Calcium 16mg the CaCo3 will be

RAM of CaCO3 is 100

percentage weight of Ca is thus

40%

therefore 16mg rep 40%

100% will be

100/40 * 160=400mg

weight of CaCo3 is thus 400mg

% weight  of CaCo3 will be 400mg/1000g* 100%

400mg/1000000mg*100% = 0.04 %

Sep 2nd, 2015

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Sep 1st, 2015
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Sep 1st, 2015
Dec 9th, 2016
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