What is the minimum volume of 6.5M HCl required to completely dissolve 24.0g of magnesium metal?
Here is a worked example. Just follow the steps.
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To find the minimum volume of 6.5M HCl required to completely dissolve 24.0g of magnesium metal, we can use the concept of stoichiometry and balanced chemical equations.
First, we need to write the balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl):
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
From the equation, we can see that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of magnesium chloride.
Next, we need to calculate the amount of magnesium in moles. The molar mass of magnesium is 24.31 g/mol, so:
n (moles of Mg) = mass (g) / molar mass (g/mol)
n (moles of Mg) = 24.0 g / 24.31 g/mol
n (moles of Mg) ≈ 0.988 moles
According to the balanced equation, we need two moles of hydrochloric acid to react with one mole of magnesium. Therefore, we would need twice as many moles of hydrochloric acid:
n (moles of HCl) = 2 * n (moles of Mg)
n (moles of HCl) ≈ 2 * 0.988 moles
n (moles of HCl) ≈ 1.976 moles
Now, we can calculate the volume of hydrochloric acid required. The molarity (M) of the hydrochloric acid is given as 6.5M, which means it contains 6.5 moles of HCl per liter (1000mL):
volume (L) = n (moles of HCl) / molarity (M)
volume (L) = 1.976 moles / 6.5 mol/L
volume (L) ≈ 0.304 L
Therefore, the minimum volume of 6.5M HCl required to completely dissolve 24.0g of magnesium metal is approximately 0.304 liters, or 304 milliliters (mL).