How many grams of MgCl2 will be obtained when 435 mL of 0.300 M HCl react with an excess of Mg(OH)2?
1 answer:
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
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