MnO2+4HCl-->MnCl2+2H20+Cl2
a. calculate the mass of MnO2 needed to produce 25 g of Cl2
b. what mass of MnCl2 is produced when 0.091 g of Cl2 is generated?
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To solve this, we need to use stoichiometry to relate the amount of MnO2 and Cl2 produced in the reaction.
The balanced equation for the reaction is:
MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2
a. To calculate the mass of MnO2 needed to produce 25 g of Cl2:
1. Find the molar mass of Cl2 (chlorine gas):
Molar mass of Cl2 = 35.45 g/mol
2. Use the stoichiometry of the balanced equation to relate the masses of Cl2 and MnO2:
From the balanced equation, we can see that 1 mole of MnO2 produces 1 mole of Cl2.
Therefore, the molar mass of MnO2 is equal to the molar mass of Cl2: 1 mol MnO2 = 1 mol Cl2
Now, set up the calculation:
(25 g Cl2) * (1 mol Cl2 / 35.45 g Cl2) * (1 mol MnO2 / 1 mol Cl2) * (86.94 g MnO2 / 1 mol MnO2)
Calculating this, we get:
25 g Cl2 * (1 mol Cl2 / 35.45 g Cl2) * (1 mol MnO2 / 1 mol Cl2) * (86.94 g MnO2 / 1 mol MnO2) ≈ 61.84 g MnO2
Therefore, approximately 61.84 grams of MnO2 would be needed to produce 25 grams of Cl2.
b. To calculate the mass of MnCl2 produced when 0.091 g of Cl2 is generated:
1. Use the stoichiometry of the balanced equation to relate the masses of Cl2 and MnCl2:
From the balanced equation, we can see that 1 mole of MnCl2 is produced for every 1 mole of Cl2.
Therefore, the molar mass of MnCl2 is equal to the molar mass of Cl2: 1 mol MnCl2 = 1 mol Cl2.
2. Set up the calculation using the given information:
(0.091 g Cl2) * (1 mol Cl2 / 35.45 g Cl2) * (1 mol MnCl2 / 1 mol Cl2) * (125.84 g MnCl2 / 1 mol MnCl2)
Calculating this, we get:
0.091 g Cl2 * (1 mol Cl2 / 35.45 g Cl2) * (1 mol MnCl2 / 1 mol Cl2) * (125.84 g MnCl2 / 1 mol MnCl2) ≈ 0.325 g MnCl2
Therefore, approximately 0.325 grams of MnCl2 is produced when 0.091 grams of Cl2 is generated.
To calculate the mass of MnO2 needed to produce 25 g of Cl2, we need to use the balanced chemical equation and the molar masses of the compounds involved.
a. Let's break down the steps to get the answer:
Step 1: Write down the balanced chemical equation:
MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2
Step 2: Find the molar mass of MnO2 and Cl2:
Molar mass of MnO2 = molar mass of Mn + 2 * molar mass of O
= 54.94 g/mol + 2 * 16.00 g/mol
= 86.94 g/mol
Molar mass of Cl2 = 2 * molar mass of Cl
= 2 * 35.45 g/mol
= 70.90 g/mol
Step 3: Use the stoichiometric coefficient from the balanced equation to find the mole ratio between MnO2 and Cl2. In this case, it is 1:1, which means for every 1 mole of MnO2, we can produce 1 mole of Cl2.
Step 4: Convert the given mass (25 g) of Cl2 into moles using the molar mass:
Moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
= 25 g / 70.90 g/mol
= 0.3526 mol
Step 5: Since the mole ratio between MnO2 and Cl2 is 1:1, the moles of MnO2 needed will be the same as moles of Cl2.
Step 6: Finally, calculate the mass of MnO2 needed using the moles and the molar mass:
Mass of MnO2 = Moles of MnO2 * Molar mass of MnO2
= 0.3526 mol * 86.94 g/mol
= 30.46 g
Therefore, the mass of MnO2 needed to produce 25 g of Cl2 is approximately 30.46 g.
b. To find the mass of MnCl2 produced when 0.091 g of Cl2 is generated, we will follow a similar approach:
Step 1: Write down the balanced chemical equation:
MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2
Step 2: Find the molar mass of MnCl2:
Molar mass of MnCl2 = molar mass of Mn + 2 * molar mass of Cl
= 54.94 g/mol + 2 * 35.45 g/mol
= 125.84 g/mol
Step 3: Use the stoichiometric coefficient from the balanced equation to find the mole ratio between Cl2 and MnCl2. In this case, it is 1:1, which means for every 1 mole of Cl2, we can produce 1 mole of MnCl2.
Step 4: Convert the given mass (0.091 g) of Cl2 into moles using the molar mass:
Moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
= 0.091 g / 70.90 g/mol
= 0.0013 mol
Step 5: Since the mole ratio between Cl2 and MnCl2 is 1:1, the moles of MnCl2 produced will be the same as moles of Cl2.
Step 6: Finally, calculate the mass of MnCl2 produced using the moles and the molar mass:
Mass of MnCl2 = Moles of MnCl2 * Molar mass of MnCl2
= 0.0013 mol * 125.84 g/mol
= 0.164 g
Therefore, the mass of MnCl2 produced when 0.091 g of Cl2 is generated is approximately 0.164 g.