How many grams of mgcl2 will be produced from 10.5g of mg (oh) 2 and 41g of hcl?

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