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Chemistry problems for mass relationships?
Hello, I have a lot of questions that need to be done for homework, but so far, there are a couple I can't do, and I need help please.
How many grams of chlorine could be obtained by heating 70 g of manganese dioxide with an excess of hydrochloric acid?
MnO2+4HCl > MnCl2 + 2H20 + Cl2
The action of dilute nitric acid on copper is represented by the equation: 3Cu +8HNO3 > 3Cu(NO3)2 + 4H2O + 2NO
What mass of copper would be required to react completely with 250 g of nitric acid? (for this part of the question I got it right) The mass turned out to 94.6 gCu.
But I can't get the second part, What mass of cupric nitrate would be formed?
For this question I don't know how to do etheir: The burning of iron pyrites (FeS2) in air may be expressed by the following equation:
4FeS2 +11O2 > 2Fe2O3 + 8SO2
a) what mass of iron pyrites must be burned in air to produce 400 g of ferric oxide?
b) what mass of oxygen will be consumed?
c) what mass of sulfur dioxide will be produced?
And finally the last one I can't get, What mass of potassium chlorate would be required to produce 45 g of oxygen?
Thanks a lot for any help, out of like 11 questions, I can't get these 4. I have a test the day after tommorow and need to understand how to do them. THANKS!
Chemistry problems for mass relationships?
Hello, I have a lot of questions that need to be done for homework, but so far, there are a couple I can't do, and I need help please.
How many grams of chlorine could be obtained by heating 70 g of manganese dioxide with an excess of hydrochloric acid?
MnO2+4HCl > MnCl2 + 2H20 + Cl2
The action of dilute nitric acid on copper is represented by the equation: 3Cu +8HNO3 > 3Cu(NO3)2 + 4H2O + 2NO
What mass of copper would be required to react completely with 250 g of nitric acid? (for this part of the question I got it right) The mass turned out to 94.6 gCu.
But I can't get the second part, What mass of cupric nitrate would be formed?
For this question I don't know how to do etheir: The burning of iron pyrites (FeS2) in air may be expressed by the following equation:
4FeS2 +11O2 > 2Fe2O3 + 8SO2
a) what mass of iron pyrites must be burned in air to produce 400 g of ferric oxide?
b) what mass of oxygen will be consumed?
c) what mass of sulfur dioxide will be produced?
And finally the last one I can't get, What mass of potassium chlorate would be required to produce 45 g of oxygen?
Thanks a lot for any help, out of like 11 questions, I can't get these 4. I have a test the day after tommorow and need to understand how to do them. THANKS!

How many grams of chlorine could be obtained by heating 70 g of manganese dioxide with an excess of hydrochloric acid?
MnO2+4HCl > MnCl2 + 2H20 + Cl2
1. Since HCl is the excess reagent (ER), MNO2 is the limiting reagent (LR). You will need the limiting reagent as a basis for the next step.
2. Get Molar Mass of MnO2
2. Convert 70 g of MnO2 into mol.
3. Use stoichiometric ratio to get the the amount of Cl2 in mol.
4. Get Molar Mass of Cl2
5. Convert the amount of Cl2 from mol to g.
Molar Mass of MnO2 = 54.94 + (2 x 16) = 86.94 g of MnO2
70 g of MnO2 x (1 mol of MnO2 / Molar Mass of MnO2) = 0.8 mol of MnO2
0.8 mol of MnO2 x (1 mol of Cl2 / 1 mol of MnO2) = 0.8 mol of Cl2
Molar Mass of Cl2 = (2 x 35.45) = 70.9 g of Cl2
0.8 mol of Cl2 x (Molar Mass of Cl2 / 1 mol of Cl2) = 56.72 g of Cl2
or you can do the shortcut altogether:
70g MnO2 x [1 mol MnO2 / 54.94 + (2x16)](1 mol Cl2 / 1 mol MnO2)[(35.45x2) g Cl2 / 1 mol Cl2]
= 57.08 g of Cl2
MnO2+4HCl > MnCl2 + 2H20 + Cl2
1. Since HCl is the excess reagent (ER), MNO2 is the limiting reagent (LR). You will need the limiting reagent as a basis for the next step.
2. Get Molar Mass of MnO2
2. Convert 70 g of MnO2 into mol.
3. Use stoichiometric ratio to get the the amount of Cl2 in mol.
4. Get Molar Mass of Cl2
5. Convert the amount of Cl2 from mol to g.
Molar Mass of MnO2 = 54.94 + (2 x 16) = 86.94 g of MnO2
70 g of MnO2 x (1 mol of MnO2 / Molar Mass of MnO2) = 0.8 mol of MnO2
0.8 mol of MnO2 x (1 mol of Cl2 / 1 mol of MnO2) = 0.8 mol of Cl2
Molar Mass of Cl2 = (2 x 35.45) = 70.9 g of Cl2
0.8 mol of Cl2 x (Molar Mass of Cl2 / 1 mol of Cl2) = 56.72 g of Cl2
or you can do the shortcut altogether:
70g MnO2 x [1 mol MnO2 / 54.94 + (2x16)](1 mol Cl2 / 1 mol MnO2)[(35.45x2) g Cl2 / 1 mol Cl2]
= 57.08 g of Cl2