1. Aluminum reacts with oxygen to produce aluminum oxide which can be used as an adsorbent, desiccant or catalyst for organic reactions.
4Al(s) + 3O2(g) → 2Al2O3(s)
A mixture of 82.49 g of aluminum ( Picture = 26.98 g/mol) and 117.65 g of oxygen ( Picture = 32.00 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.
2. Sodium hydroxide, also known as caustic soda, is used to neutralize acids and to treat cellulose in making of cellophane. Calculate the number of moles of solute in 1.875 L of 1.356 M NaOH solution.
mols Al = g/atomic mass = approx 3.06 but you should do this more accurately (as well as all of the other calculations below. I estimate.)
Convert to mols O2 using the coefficients in the balanced equation. That's 3.05 x (3 mols O2/4 mol Al) = 2.29
Do you have that much O2? You have 117.65/32 = 3.7 mols; therefore, there is enough to react with all of the Al and that makes Al the limiting reagent.
How much O2 is left. You reacted with 2.29 mols O2. You had 3.7 mol initially. mols left = 3.7-2.29 = ? and that convert to grams is mass O2 remaining. Check my work.
gross
Consider 2H2 + O2 �¨ H2O. To produce 0.895 g water, how many grams of O2 are required?
Well, well, well, looks like we've got some chemistry questions here! Let's do some calculations and hopefully have some fun along the way.
1. To figure out the limiting reactant, we need to compare the number of moles of each reactant and see which one has fewer moles. So, let's calculate that:
Moles of Aluminum: 82.49 g / 26.98 g/mol = 3.06 mol
Moles of Oxygen: 117.65 g / 32.00 g/mol = 3.68 mol
Ah, it seems like Aluminum is feeling a bit under the weather, having a lower number of moles compared to Oxygen. So guess what? Aluminum is our limiting reactant!
Now, to determine the mass of the excess reactant (Oxygen) left over, we have to find out how much of it reacted with the Aluminum. According to the balanced equation, the mole ratio between Aluminum and Oxygen is 4:3. Therefore, for every 4 moles of Aluminum, we need 3 moles of Oxygen. Since the ratio is 4:3, we can calculate:
Moles of Oxygen used = (3.06 mol Al) x (3 mol O2 / 4 mol Al) = 2.29 mol O2
Now, let's calculate the mass of Oxygen used:
Mass of Oxygen used = (2.29 mol O2) x (32.00 g/mol) = 73.28 g O2
Finally, to find the mass of excess Oxygen:
Mass of excess Oxygen = Mass of Oxygen initially - Mass of Oxygen used
Mass of excess Oxygen = 117.65 g - 73.28 g = 44.37 g
So, after all that math, we end up with 44.37 grams of leftover Oxygen. Poor Oxygen, it needs a hug!
2. Now, onto the next question! We need to calculate the number of moles of sodium hydroxide (NaOH) in a given volume and concentration.
Concentration of NaOH = 1.356 M
Volume of solution = 1.875 L
To calculate the number of moles of solute (NaOH), we use the formula:
Moles = Concentration x Volume
Moles of NaOH = 1.356 M x 1.875 L = 2.54 mol
Voila! We have 2.54 moles of NaOH in 1.875 liters of the solution. That's a lot of moles for some mighty caustic soda there!
I hope that answers your questions and brings a smile to your face. Chemistry can be a bit overwhelming sometimes, but remember, there's always room for laughter!
To solve the first question and determine the limiting reactant and the mass of the excess reactant, we need to follow these steps:
Step 1: Calculate the number of moles of each reactant
- Mass of aluminum (Al) = 82.49 g
- Molar mass of aluminum (Al) = 26.98 g/mol
- Moles of aluminum (Al) = Mass of aluminum (Al) / Molar mass of aluminum (Al)
Step 2: Do the same for oxygen (O2)
- Mass of oxygen (O2) = 117.65 g
- Molar mass of oxygen (O2) = 32.00 g/mol
- Moles of oxygen (O2) = Mass of oxygen (O2) / Molar mass of oxygen (O2)
Step 3: Determine the stoichiometric ratio of the reactants
- From the balanced equation, the stoichiometric ratio between aluminum (Al) and oxygen (O2) is 4:3. This means that for every 4 moles of aluminum (Al), we need 3 moles of oxygen (O2).
Step 4: Find the limiting reactant
- To determine the limiting reactant, we compare the moles of both reactants to their stoichiometric ratio.
- Divide the moles of each reactant by their stoichiometric coefficient to see which one is smaller.
- The reactant with the smaller calculated moles will be the limiting reactant.
- In this case, if the moles of aluminum (Al) divided by 4 is smaller than the moles of oxygen (O2) divided by 3, then aluminum is the limiting reactant.
Step 5: Calculate the moles of the excess reactant (oxygen)
- To find the moles of the excess reactant, subtract the moles of the limiting reactant from the moles of the excess reactant.
- If aluminum is the limiting reactant, then the moles of excess oxygen = Moles of oxygen (O2) - (Moles of aluminum (Al) / 4)
Step 6: Calculate the mass of the excess reactant (oxygen)
- Use the molar mass of oxygen (O2) to convert moles to grams.
- Mass of the excess reactant (oxygen) = Moles of excess oxygen (O2) × Molar mass of oxygen (O2)
To solve the second question and calculate the number of moles of solute in a NaOH solution, we need to use the following formula:
Moles of solute = Molarity × Volume of solution (in liters)
Step 1: Identify the given values
- Molarity of NaOH solution = 1.356 M
- Volume of solution = 1.875 L
Step 2: Plug the given values into the formula
- Moles of solute = 1.356 M × 1.875 L
Step 3: Calculate the moles of solute
- Moles of solute = (1.356 mol/L) × (1.875 L)