306 grams of an unknown is composed of two substances in unknown quantities: MgS

Question

306 grams of an unknown is composed of two substances in unknown quantities: MgSO4x7D2O and CaCO3. Use 153 grams in this reaction MgSO4 x 7D2O yield Mg SO4 x 4D2O plus 3D2O crucible mass is 100 grams and crucible mass plus product is 233 grams The other reaction is CaCO plus 2HCL yields CaCl2 plus h2o. 17.1 of CO2 is produced and the pressure is 760 torr the temp is 300K. What percent of unknown is Magnesium Sulfate heptadeuterate by mass and by moles? What percent of unknown is calcium carbonate by mass and by moles? How many moles each do you have? Does your unknown have any impurities?

Explanation / Answer

MgSO4 x 7D2O ----> Mg SO4 x 4D2O + 3D2O

mass of product = 233-100 = 133 grams

No of moles of product = 133/260.366 = 0.5115 mole

Molar mass of MgSO4.7D2O = 260.366 g/mol

so that,

No of moles MgSO4 x 7D2O involves in reaction = 0.5115 mole

in 153 grams of sample 133 grams of MgSO4.7D2O is present.

therefore

in 306 grams sample mass of MgSO4.7D2O = 306*133/153 = 266 grams

CaCO3 + 2HCL ---> CaCl2 + H2O + CO2(g)

no of moles of CO2 produced = 17.1/44 = 0.39 mole

from equation

1 mole CaCO3 = 1 mole co2

No of moles of CaCO3 reacted   = 0.39 mole

mass of CaCO3 = 0.39*100.0869 = 39.034 grams

Molar mass of caco3 = 100.0869 g/mol

Total mass of MgSO4 x 7D2O + CaCO3 present in sample = 266+39.04 = 305.04 grams

actual mass of sample = 306 grams

impurities = 306-305.04 = 0.96 grams

percent of calcium carbonate by mass = 39.04/306*100 = 12.76% by mass

percent of calcium carbonate by moles = 0.39/(0.39+0.511)*100 = 43.28%

percent of MgSO4 x 7D2O by mass = 266/306*100 = 86.92 %

percent of MgSO4 x 7D2O by moles = 0.511/(0.39+0.511)*100 = 56.71 %

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