Question

How many g of MgCO3(s) are needed to make 1.2 L of 1.5 M MgCl2(aq) solution? Please balance the equation before solving the problem.

Answer 1

the grams of MgCO3 that are needed to make 1.2 l of 1.5M MgCl2 solution is calculated as below
MgCO3 +2HCl = MgCl2 +CO2 +H2O

find the moles of MgCl2 produced

moles= molarity x volume

= 1.2 x 1.5 = 1.8 moles
by use of mole ratio between MgCO3 to MgCl2 which is 1:1 the moles of MgCo3 is therefore = 1.8moles

mass of MgCo3 =moles x molar mass
= 1.8moles x 84.3 g/mol= 151.74 grams of MgCO3