Question

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)

Answer 1

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

Answer 2

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 %