When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 4.8 M is required to react completely with 2.60 g of magnesium?
Balance the equation;
2HCl + Mg>>>Mg(Cl)2 + H2
so it takes two moles of HCL per mole of Mg.
how many moles does one have in 2.60gMg
ans: 2.60/atomicmassMg
Now take twice of that for moles of acid.
Molarityacid=molesacid/volume
volume=molesacid/4.8
To determine the volume of 4.8 M hydrochloric acid required to react completely with 2.60 g of magnesium, we need to follow a series of steps.
Step 1: Write and balance the chemical equation for the reaction between hydrochloric acid (HCl) and magnesium (Mg):
Mg + 2 HCl → MgCl₂ + H₂
Step 2: Calculate the moles of magnesium (Mg) using its molar mass. The molar mass of Mg is 24.31 g/mol.
moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 2.60 g / 24.31 g/mol = 0.107 mol
Step 3: Use the stoichiometry of the balanced chemical equation to find the moles of hydrochloric acid (HCl) required.
From the balanced equation, we see that 1 mole of Mg reacts with 2 moles of HCl. Therefore, the moles of HCl needed is twice the moles of Mg.
moles of HCl = 2 * moles of Mg
moles of HCl = 2 * 0.107 mol = 0.214 mol
Step 4: Calculate the volume of hydrochloric acid using the given concentration (4.8 M). Remember, M represents moles per liter (mol/L).
volume of HCl = moles of HCl / concentration of HCl
volume of HCl = 0.214 mol / 4.8 mol/L
Step 5: Convert the concentration to moles per milliliter (mol/mL).
1 L = 1000 mL
So, the concentration becomes 4.8 mol/L * (1 L / 1000 mL) = 0.0048 mol/mL
Step 6: Calculate the volume of hydrochloric acid in milliliters (mL).
volume of HCl = 0.214 mol / 0.0048 mol/mL
volume of HCl = 44.58 mL
Therefore, approximately 44.58 mL of 4.8 M hydrochloric acid is required to react completely with 2.60 g of magnesium.