On the same subject as before with the limiting/excess reagent stuff -- can someone please check my answer to make sure I'm doing it right? Thanks sooooo much!! :DD
3. Mg + 2HCl ==> MgCl2 + H2 -- What mass of hydrogen at STP is produced from the reaction of 50 g of Mg and the equivalent of 75 g of HCl?
==> I got about 2.08 g H2.
4. How much of the excess reagent in Problem 3 is left over?
==> I got about 12 g Mg.
Oops, I mean I got 49 g Mg for number 4.
To determine if your answers are correct, we can go through the steps of solving the problem together.
First, let's address problem 3:
Given:
- 50 g Mg
- The equivalent of 75 g HCl
To find the mass of hydrogen produced, we need to determine which reactant is the limiting reagent. We can do this by comparing the moles of Mg and HCl.
1. Convert the mass of Mg to moles:
- Calculate the molar mass of Mg (24.31 g/mol)
- Divide the given mass (50 g) by the molar mass to find the number of moles of Mg
2. Convert the mass of HCl to moles:
- Calculate the molar mass of HCl (36.46 g/mol)
- Divide the given mass (75 g) by the molar mass to find the number of moles of HCl
3. Determine the mole ratio based on the balanced chemical equation:
- The coefficient of Mg is 1, while the coefficient of HCl is 2
- This means that the mole ratio of Mg to HCl is 1:2
4. Compare the moles:
- Divide the number of moles of Mg by 1 (since it has a coefficient of 1)
- Divide the number of moles of HCl by 2 (since it has a coefficient of 2)
- Whichever value is smaller represents the limiting reagent
5. Calculate the theoretical yield of hydrogen:
- Based on the mole ratio, determine how many moles of hydrogen would be produced from the limiting reagent
- Multiply the number of moles of hydrogen by its molar mass (2.016 g/mol) to find the mass of hydrogen produced
Now, let's apply the steps above to problem 3:
1. Moles of Mg:
- Molar mass of Mg = 24.31 g/mol
- Moles of Mg = 50 g / 24.31 g/mol ≈ 2.06 mol
2. Moles of HCl:
- Molar mass of HCl = 36.46 g/mol
- Moles of HCl = 75 g / 36.46 g/mol ≈ 2.06 mol
3. Since both substances have the same number of moles, neither is in excess. Therefore, we can proceed with calculating the mass of hydrogen.
4. Moles of H2:
- Since the mole ratio is 1:2 (Mg:HCl), we have 2.06 mol HCl * (1 mol H2 / 2 mol HCl) = 1.03 mol H2
5. Mass of H2:
- Molar mass of H2 = 2.016 g/mol
- Mass of H2 = 1.03 mol * 2.016 g/mol ≈ 2.08 g H2
Therefore, your answer of approximately 2.08 g H2 for problem 3 is correct.
Now, let's tackle problem 4:
Given:
- Excess reagent: Mg
- Theoretical yield of H2: 2.08 g
To determine the excess reagent remaining, we need to calculate the moles of Mg reacting with the limiting reagent (HCl) and subtract it from the initial moles of Mg.
1. Moles of Mg reacting with HCl:
- Since the mole ratio is 1:2 (Mg:HCl), 1.03 mol H2 * (1 mol Mg / 1 mol H2) = 1.03 mol Mg
2. Moles of excess Mg remaining:
- Initial moles of Mg = 50 g Mg / 24.31 g/mol ≈ 2.06 mol Mg
- Moles of excess Mg remaining = Initial moles of Mg - Moles of Mg reacting = 2.06 mol - 1.03 mol = 1.03 mol
3. Mass of excess Mg remaining:
- Mass of excess Mg remaining = Moles of excess Mg * 24.31 g/mol ≈ 25.1 g Mg
Therefore, your answer of approximately 12 g Mg for problem 4 is not correct. Based on the calculations, the correct answer is approximately 25.1 g Mg.
I hope this explanation helps! Let me know if you have any further questions.