If 84.1 g of sodium hydroxide are reacted with 51.0 g of aluminum in the following equation:
2NaOH + 2Al + 2H2O --> 2NaAlO2 + 3H2
a) what mass of excess reactant remains?
b) what mass of water would react?
Note that NaOH has a molar mass of 40 g/mol, and Al has a molar mass of 27 g/mol. If you're not sure how to get the molar mass, you'll need a periodic table, and add the individual masses of each element present in the chemical formula.
To solve this, first we need to get the moles of each reactant. To get them, we just divide the given mass by the molar mass:
NaOH: 84.1 g / 40 g/mol = 2.1025 mol NaOH
Al: 51 g / 27 g/mol = 1.8889 mol Al
Then from the balanced reaction, we make a mole ratio of reactants:
2 mol NaOH / 2 mol Al or
2 mol Al / 2 mol NaOH
From the moles of reactant we calculated above, we get the corresponding moles of the other reactant using the mole ratio of reactants:
2.1025 mol NaOH * (2 mol Al / 2 mol NaOH) = 2.1025 mol Al
1.8889 mol Al * (2 mol NaOH / 2 mol Al) = 1.8889 mol NaOH
From here we can see that the limiting reactant is Al, because in order to completely react the 2.1025 mol NaOH, it needs 2.1025 mol Al, but in the given we only have 1.8889 mol Al.
To get the mass of excess reactant (NaOH), we subtract the moles from above,
2.1025 - 1.8889 = 0.2136 mol NaOH
mass,NaOH = 0.2136 mol * 40 g/mol = 8.544 g
To get the mass of water, same thing we'll do. We use the moles of the limiting reactant and mole ratio of reactant to get the moles of water:
1.8889 mol Al * (2 mol H2O / 2 mol Al) = 1.8889 mol H2O
Since water is 18 g/mol, just multiply the this with the moles to get the mass.
hope this helps~ `u`
Thank you!!
To determine the mass of excess reactant remaining and the mass of water that would react, we need to first calculate the amount of reactant used in the reaction.
Step 1: Convert the given masses of sodium hydroxide and aluminum into moles using their respective molar masses.
- Molar mass of NaOH = 22.99 g/mol (sodium) + 16.00 g/mol (oxygen) + 1.01 g/mol (hydrogen) = 39.99 g/mol
- Number of moles of NaOH = 84.1 g / 39.99 g/mol = 2.10 mol
- Molar mass of aluminum = 26.98 g/mol
- Number of moles of aluminum = 51.0 g / 26.98 g/mol = 1.89 mol
Step 2: Determine the limiting reactant by comparing the mole ratios of NaOH and Al in the balanced chemical equation.
- From the balanced equation, we can see that the ratio of NaOH to Al is 2:2 or 1:1.
- Since the number of moles of NaOH (2.10) is greater than the number of moles of aluminum (1.89), aluminum is the limiting reactant.
Step 3: Calculate the amount of excess reactant remaining. In this case, NaOH is the excess reactant.
- To do this, we need to determine the amount of NaOH that reacts with the available moles of aluminum.
- According to the balanced equation, 2 moles of NaOH react with 2 moles of Al. Hence, 1 mole of NaOH reacts with 1 mole of Al.
- Therefore, the amount of NaOH that reacts with 1.89 moles of Al is 1.89 moles.
- The initial amount of NaOH was 2.10 moles, so the mass of excess NaOH remaining is:
= (2.10 mol - 1.89 mol) x 39.99 g/mol
= 8.36 g
a) The mass of excess NaOH remaining is 8.36 g.
Step 4: Calculate the amount of water that reacts.
- According to the balanced equation, 2 moles of NaOH react with 2 moles of H2O.
- Therefore, the moles of H2O is the same as the moles of NaOH that reacted.
- The moles of H2O that react with 1.89 moles of Al is 1.89 moles.
b) The mass of water that would react is:
= 1.89 mol x 18.02 g/mol
= 34.05 g
b) The mass of water that would react is 34.05 g.
To find the mass of excess reactant remaining, we need to determine which reactant is in excess and which is limiting.
First, we'll calculate the number of moles for each reactant using their molar masses. The molar mass of NaOH is 40.00 g/mol, and the molar mass of Al is 26.98 g/mol.
a) Mass of NaOH (in g) = 84.1 g
Molar mass of NaOH = 40.00 g/mol
Moles of NaOH = mass/molar mass = 84.1 g / 40.00 g/mol ≈ 2.1025 mol
Mass of Al (in g) = 51.0 g
Molar mass of Al = 26.98 g/mol
Moles of Al = mass/molar mass = 51.0 g / 26.98 g/mol ≈ 1.8896 mol
According to the balanced equation, the molar ratio between NaOH and Al is 2:2. So, the moles of both are equal. This means NaOH is not in excess, and Al is in excess. We'll use Al to determine the mass of the excess reactant remaining.
To find the mass of the excess reactant remaining, we'll use the stoichiometry of the reaction.
Moles of NaOH used = moles of Al used ≈ 1.8896 mol
Now, we need to find the mass of NaOH used in this reaction:
Mass of NaOH used = moles of NaOH × molar mass of NaOH
≈ 1.8896 mol × 40.00 g/mol
≈ 75.584 g
The mass of excess reactant remaining is the difference between the initial mass and the used mass of Al:
Mass of excess reactant remaining = Initial mass of Al - Mass of Al used
= 51.0 g - 75.584 g
≈ -24.584 g
From the calculation, we find that the mass of the excess reactant remaining is approximately -24.584 g. Since this value is negative, it means that the excess reactant (Al) has been completely consumed, and no excess reactant remains.
b) To calculate the mass of water that would react, we need to determine the limiting reactant first. Since Al is the excess reactant, we'll use the moles of NaOH to do the calculations.
According to the balanced equation, 2 moles of NaOH react with 2 moles of Al and produce 3 moles of H2O.
Moles of H2O produced = (2 moles of NaOH) × (3 moles of H2O / 2 moles of NaOH)
= 3 moles of H2O
To find the mass of water, we'll use the molar mass of water, which is 18.02 g/mol.
Mass of water produced = moles of H2O produced × molar mass of H2O
= 3 moles × 18.02 g/mol
= 54.06 g
Therefore, the mass of water that would react is approximately 54.06 g.