Answer:
33.33% = 33%
Explanation:
MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)
1 mole of MCO3 will produce → 1 mole of CO2
We need to get the 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 %
