In a reaction, 25.0 grams of magnesium are added to a beaker containing 75.0 mL of 2.00 M nitric acid, HNO3. Which reactant is limiting? Which is in excess?
How many moles of Mg do you have? (25/23.5 ?)
how many moles of H do you have? (.075*2)
The reaction:
Mg+2HNO3>>H2+Mg(NO3)2
so is the nitric acid moles twice the Mg? If moles Nitric acid is >2*moles Mg, then Mg is limiting.
using the G.U.E.S.S. method, how do I figure out this equation?!!!
You have the equation in the problem.
To determine which reactant is limiting and which is in excess, we need to perform a stoichiometric calculation.
Step 1: Write the balanced equation for the reaction:
Mg + 2HNO3 -> Mg(NO3)2 + H2
Step 2: Calculate the moles of magnesium (Mg):
Molar mass of Mg = 24.3 g/mol
Moles of Mg = mass / molar mass
Moles of Mg = 25.0 g / 24.3 g/mol ≈ 1.03 mol
Step 3: Calculate the moles of nitric acid (HNO3):
Molar mass of HNO3 = 63.0 g/mol
Moles of HNO3 = concentration × volume (in liters)
Moles of HNO3 = 2.00 mol/L × 0.075 L ≈ 0.150 mol
Step 4: Determine the stoichiometry of the reaction:
From the balanced equation, we see that 1 mole of Mg reacts with 2 moles of HNO3.
Step 5: Compare the moles of each reactant:
Considering the stoichiometry, we have:
Moles of Mg: 1.03 mol
Moles of HNO3: 0.150 mol
Since the stoichiometry of the reaction between Mg and HNO3 is 1:2, we can see that there is an excess of nitric acid. This means that nitric acid is in excess, while magnesium is the limiting reactant.