How many grams of mgcl2 will be obtained when 435 ml of 0.300 m hcl react with an excess of mg(oh)2?

The grams of Mg(OH)2 produced is calculated as below

calculate the moles of HCl produced =molarity  xvolume/1000

= 0.3 x 435/1000= 0.1305 moles

write the equation for reaction
Mg(OH)2 +2HCl = MgCl2 + 2H2O

by use of mole ratio between HCl :MgCl2  which is 2 :1  the  moles of HCl =  0.1305 x1/2 =0.0653  moles of MgCl2

mass of MgCl2 = moles x molar mass

= 0.0653mol x95 g/mol = 6.204  grams of MgCl2