<span>The Answer </span>should be<span> .074 M</span>
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 %
