cheekmartleyRussel
Answered

The mineral magnesite contains magnesium carbonate, MgCO3 (molar mass= 84 g/mol), and other impurities. When a 1.26-g sample of magnesite was dissolved in hydrochloric acid, 0.22 g of CO2 was generated. If the magnesite contained no carbonate other than MgCO3, what was the percent MgCO3 by mass in the magnesite? The reaction is as follows:


MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)

Verified Answer :

meerkat18

we are given with the equation MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g). we shall convert CO2 to number of grams of MgCO3. Via stiochiometry, 0.22 g CO2 is equal to 0.42 gram MgCO3. In this case, the purity of MgCO3 in the sample is 0.42 g / 1.26 g or equal to 33.33 percent

superman1987

The answer is 33.33 %


The explanation:


According to the reaction equation:


MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)


we can see that 1 mole of MCO3 will produce → 1 mole of CO2


-Now we need o get number of mole of CO2:


and when we have 0.22 g of CO2, so number of mole = mass / molar mass


moles = 0.22 g / 44 g/mol = 0.005 mole


∴ moles of Mg = moles of CO2 = 0.005 mole


∴ mass of Mg = moles molar mass


= 0.005 84 /mol = 0.42 g


∴ Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100


= 33.33 %



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