Answer:
1.48 M
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
Mg + H2SO4 —> MgSO4 + H2
Step 2:
Determination of the number of mole of Mg in 80.0 mL of 0.200 M Mg solution. This is illustrated below:
Molarity of Mg = 0.200 M
Volume of solution = 80 mL = 80/1000 = 0.08L
Mole of Mg =?
Molarity = mole /Volume
0.2 = mole /0.08
Mole = 0.2 x 0.08
Mole of Mg = 0.016 mole.
Step 3:
Determination of the number of mole of H2SO4 that reacted. This is illustrated below:
Mg + H2SO4 —> MgSO4 + H2
From the balanced equation above,
1 mole of Mg reacted with 1 mole of H2SO4.
Therefore, 0.016 mole of Mg will also react with 0.016 mole of H2SO4.
Step 4:
Determination of the concentration of the acid.
Mole of H2SO4 = 0.016 mole.
Volume of acid solution = 10.8 mL = 10.8/1000 = 0.0108 L
Molarity =?
Molarity = mole /Volume
Molarity = 0.016/0.0108
Molarity of the acid = 1.48 M
Therefore, the concentration of acid is 1.48 M
