Can anyone help me? Thank you in advance!
1.) Given: 2H2O2=2H2O+O2
a.) What mass of water is produced when 5.0 mol O2 is produced?
2.) Given: 3Fe+4H2O=4H2+Fe3O4
a.) If 625g of Fe3O4 is produced on the reaction how many moles of hydrogen are produced at the same time?
b.) How many moles of iron would be needed to generate 27g of hydrogen?
3.) Given: MnO2+4HCl2=MnCl2+2H20+Cl2
a.) Calculate the mass of MnO2 needed to produced 25 Og of Cl2
b.) What mass of MnCl2 is produced when 0.091g of Cl2 is generated?
Answers & Comments
HELLO FOLK!
1 a.) To solve this problem, we first need to determine the number of moles of water produced when 5.0 mol of O2 is produced. From the balanced chemical equation, we know that 2 moles of H2O are produced for every mole of O2. Therefore, the number of moles of H2O produced is:
(5.0 mol O2) x (2 mol H2O/1 mol O2) = 10.0 mol H2O
To calculate the mass of H2O produced, we use the molar mass of water, which is 18.015 g/mol:
(10.0 mol H2O) x (18.015 g/mol H2O) = 180.15 g H2O
Therefore, 180.15 g of water is produced when 5.0 mol of O2 is produced.
2 a.) To solve this problem, we need to use the balanced chemical equation to determine the mole ratio between Fe3O4 and H2:
3Fe + 4H2O → 4H2 + Fe3O4
From the equation, we see that 3 moles of Fe are needed to produce 4 moles of H2. Therefore, the number of moles of H2 produced can be calculated as follows:
(625 g Fe3O4) x (1 mol Fe3O4/231.53 g Fe3O4) x (4 mol H2/3 mol Fe3O4) = 4.34 mol H2
Therefore, 4.34 moles of H2 are produced when 625 g of Fe3O4 is produced.
2 b.) To solve this problem, we need to use the mole ratio between Fe and H2 from the balanced chemical equation. We can start by calculating the number of moles of H2 produced from 27 g:
27 g H2 x (1 mol H2/2.016 g H2) = 13.42 mol H2
From the balanced equation, we see that 3 moles of Fe are needed to produce 4 moles of H2. Therefore, the number of moles of Fe needed can be calculated as follows:
13.42 mol H2 x (3 mol Fe/4 mol H2) = 10.07 mol Fe
Therefore, 10.07 moles of Fe would be needed to generate 27 g of H2.
3 a.) To solve this problem, we need to use the mole ratio between MnO2 and Cl2 from the balanced chemical equation:
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
From the equation, we see that 1 mole of MnO2 produces 1 mole of Cl2. Therefore, the number of moles of MnO2 needed to produce 25 g of Cl2 can be calculated as follows:
25 g Cl2 x (1 mol Cl2/70.906 g Cl2) x (1 mol MnO2/1 mol Cl2) = 0.352 mol MnO2
Therefore, 0.352 mol MnO2 is needed to produce 25 g of Cl2.
3 b.) To solve this problem, we need to use the mole ratio between MnCl2 and Cl2 from the balanced chemical equation. We can start by calculating the number of moles of Cl2 from 0.091 g:
0.091 g Cl2 x (1 mol Cl2/70.906 g Cl2) = 0.00128 mol Cl2
From the equation, we see that 1 mole of MnCl2 is produced for every 1 mole of Cl2. Therefore, the number of moles of MnCl2 produced can be calculated as follows:
0.00128 mol Cl2 x (1 mol MnCl2/1 mol Cl2) x (125.84 g MnCl2/1 mol MnCl2) = 0.161 g MnCl2
Therefore, 0.161 g of MnCl2 is produced when 0.091 g of Cl2 is generated.