How many g of mgco3(s) are needed to make 1.2 l of 1.5 m mgcl2(aq) solution? balance the equation before solving the problem.

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