When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 5.0 M HCI is required to react completely with 3.00 g of magnesium?
The Chemical reaction is:
Mg + 2HCl --> MgCl2 + H2
1) Divide the grams of Mg by the atomic mass of Mg to get moles of Mg.
2) Moles of HCl = (2)(moles of Mg)
3) Moles of HCl = (V)(5.0 mol/L)
or
Volume of HCl = (mol. HCl) / (5.0 mol/L)
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NOTE: The volume of HCl will be in liters
If 5.00 g of manganese (IV) oxide reacts with 25.0 mL of 6.00 M hydrochloric acid, what is the calculated mass of chlorine gas produced?
To determine the volume of 5.0 M hydrochloric acid (HCI) required to react completely with 3.00 g of magnesium, we can use a stoichiometry calculation.
First, we need to write a balanced equation for the reaction between hydrochloric acid and magnesium:
2 HCI + Mg → H2 + MgCl2
From the balanced equation, we can see that 2 moles of hydrochloric acid react with 1 mole of magnesium to produce 1 mole of hydrogen gas.
Step 1: Convert the mass of magnesium to moles.
Using the molar mass of magnesium (24.31 g/mol), we can calculate the number of moles of magnesium:
3.00 g Mg × (1 mol Mg / 24.31 g Mg) = 0.1233 mol Mg
Step 2: Determine the number of moles of hydrochloric acid required.
Since the stoichiometric ratio between hydrochloric acid and magnesium is 2:1, we multiply the number of moles of magnesium by 2:
0.1233 mol Mg × (2 mol HCI / 1 mol Mg) = 0.2466 mol HCI
Step 3: Calculate the volume of 5.0 M hydrochloric acid required.
The concentration of 5.0 M means there are 5.0 moles of hydrochloric acid in 1 liter (1000 mL) of solution. Therefore, to calculate the volume of 0.2466 mol of hydrochloric acid:
Volume (L) = Moles of HCI / Concentration (M)
Volume (L) = 0.2466 mol HCI / 5.0 mol/L = 0.04932 L
To convert the volume to milliliters (mL), multiply by 1000:
Volume (mL) = 0.04932 L × 1000 = 49.32 mL
Therefore, approximately 49.32 mL of 5.0 M hydrochloric acid is required to react completely with 3.00 g of magnesium.
To answer this question, we need to use stoichiometry and the balanced chemical equation for the reaction between hydrochloric acid (HCl) and magnesium (Mg).
First, let's write the balanced equation for the reaction:
2HCl + Mg -> MgCl2 + H2
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of Mg. Now, let's calculate the number of moles of magnesium (Mg) we have:
Number of moles = Mass / molar mass
The molar mass of magnesium (Mg) is 24.31 g/mol, so:
Number of moles of Mg = 3.00 g / 24.31 g/mol = 0.1235 mol (rounded to 4 decimal places)
Since the stoichiometric ratio between HCl and Mg is 2:1, the number of moles of HCl required to react with 0.1235 moles of Mg is double that amount:
Number of moles of HCl = 2 * 0.1235 mol = 0.247 mol
Finally, we can determine the volume of 5.0 M HCl needed. The molarity (M) of a solution is defined as the number of moles of solute per liter of solution.
Volume of HCl = Number of moles / Molarity
Volume of HCl = 0.247 mol / 5.0 mol/L = 0.0494 L
To convert the volume from liters to milliliters, multiply by 1000:
Volume of HCl = 0.0494 L * 1000 mL/L = 49.4 mL
Therefore, approximately 49.4 mL of 5.0 M HCl is required to react completely with 3.00 g of magnesium.