Question

How many grams of MgCl2 (molar mass = 95.20 g/mol) will be formed from 25.6 mL of a 0.100 M HCL solution reacting with excess magnesium metal? (The other product is hydrogen gas.)

Answer 1

the balanced equation for the above reaction is as follows;
Mg + 2HCl ---> MgCl₂ + H₂
stoichiometry of HCl to MgCl₂ is 2:1
we have been told that Mg is in excess therefore HCl is the limiting reactant 
number of HCl moles reacted - 0.100 mol/L x 0.0256 L = 0.00256 mol
according to molar ratio, number of MgCl₂ moles formed - 0.00256/2 
Therefore number of MgCl₂ moles formed - 0.00128 mol
mass of MgCl formed - 0.00128 mol x 95.20 g/mol = 0.122 g

Answer 2

Answer:

0.1219 g

Explanation:

mole = mass/molar mass = molarity x volume

From the balanced equation of reaction:

$Mg + 2HCl ---> MgCl_2 + H_2$

1 mole of MgCl2 is formed from 2 moles of HCl

x mole of MgCl2 is formed from 0.1 x 0.0256 moles HCl

x mole of MgCl2 = 1 x 0.1 x 0.0256/2

                               = 0.00128

0.00128 moles of MgCl2 is formed from 0.00256 moles HCl

Hence,

mass of MgCl2 = moles x molar mass

                          = 0.00128 x 95.20

                            = 0.1219 g

Therefore, 0.1219 g of MgCl2 will be formed from 25.6 mL of a 0.100 M HCL solution reacting with excess magnesium metal.