The number of grams of MgCl2 that will be produced from 10.5 g Mg(OH)2 and 41 of HCl is 17.154 grams
calculation
write the equation for reaction
Mg(OH)2 + HCl → MgCl2 +2H2O
find the moles of each compound used
moles=mass /molar mass
moles of Mg(OH)2 = 10.5g/ 58.3 g/mol = 0.180 moles
moles of HCl = 41 g/ 36.5g/mol = 1.123 moles
by use of of mole ratio between Mg(OH)2 to MgCl2 which is 1:1 moles of MgCl2 is also 0.180 moles
by use of mole ratio between HCl to MgCl2 which is 2:1 the moles of MgCl2 = 1.123 x1/2 = 0.5615 moles
HCl was in excess while Mg(OH)2 was the limiting reagent, therefore moles of MgCl2 = 0.180 moles
mass of MgCl2= moles MgCl2 x molar mass of MgCl2 ( 24.3 + 35.5 x2=95.3)
mass = 0.180 x 95.3 = 17.154 grams
