Magnesium metal reacts with hydrochloric acid according to the following equation: Mg(s) + 2HCl (aq) ---> MgCl2(aq) + H2. What volume of 0.0400 M HCl(aq) is required to react with 6.08 grams of Mg(s)?
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The equation of the chemical reaction between magnesium and hydrochloric acid is: Mg(s) + 2HCl (aq) ---> MgCl2 (aq) + H2
From the equation it is seen that 2 moles of hydrochloric acid are needed to react with one mole of magnesium.
The molar mass of magnesium is 24.30 g/mole. 6.08 g of magnesium is equivalent to 6.08/24.30 or approximately 0.25 moles. 0.25*2 = 0.5 mole of hydrochloric acid is required to react with the magnesium. As the concentration of hydrochloric acid is 0.04 M or 0.04 mole/liter, the volume of acid required is 2*0.25/0.04 = 12.5 liters.
The volume of 0.04 M hydrochloric acid required for the given reaction with 6.08 g of magnesium is 12.5 liters.
The balanced reaction is
Mg(s) + 2HCl (aq) ---> MgCl2(aq) + H2
here we can see that the molar artio berween Mg and HCl is 1:2. That is 1mole of Mg requries 2 mol of HCl.
Mole = mass/molar mass
Moles of Mg = 6.08g/(24.31 g/mol) [molar mass of Mg is 24.31 g/mol]
Moles of Mg = 0.25 mol Mg
So moles of HCl requried would be
0.25 Mg * (2 Mol HCl/1 mol Mg)
0.50 mol HCl
we know that
Moraity = moles/volume
volume = moles/molarity
Volume of HCl = 0.50 mol /0.0400 M = 12.5 L
So 12.5 Liter of 0.040M HCl is required.
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