Using the balanced chemical equation:
2H2(g) + O2(g) → 2H2O(g)
we can see that the stoichiometry of the reaction is 2:1, meaning that for every 2 moles of hydrogen reacted, 1 mole of oxygen is reacted.
First, we convert the given volumes of hydrogen and oxygen to moles using the ideal gas law:
n(H2) = PV/RT = (80 cm^3) (1 atm) / (0.0821 L*atm/mol*K) (273 K)
n(H2) = 3.07 x 10^-3 mol
n(O2) = PV/RT = (25 cm^3) (1 atm) / (0.0821 L*atm/mol*K) (273 K)
n(O2) = 9.57 x 10^-4 mol
Since the stoichiometry of the reaction is 2:1, the limiting reactant is oxygen. Therefore, all of the oxygen will be consumed and 4.78 x 10^-4 mol of hydrogen will react.
Now we can use the ideal gas law again to calculate the volume of residual gases:
V = nRT/P
For hydrogen:
V = (3.07 x 10^-3 mol) (0.0821 L*atm/mol*K) (273 K) / (1 atm)
V = 0.0661 L = 66.1 cm^3
For water vapor (product):
n(H2O) = 4.78 x 10^-4 mol (from stoichiometry)
V = (4.78 x 10^-4 mol) (0.0821 L*atm/mol*K) (273 K) / (1 atm)
V = 0.0102 L = 10.2 cm^3
Therefore, the volume of residual gas is:
Vresidual = Vtotal - Vreacted
Vresidual = (80 cm^3 + 25 cm^3) - (66.1 cm^3 + 10.2 cm^3)
Vresidual = 28.7 cm^3
In reaction 2H2(g)+ow(g) to 2H2O(g), 80 centimeter square of hydrogen was reacted with 25centmetercube of oxygen, calculate the volume of residual gasses
5 answers
15.0(g)of potassium trioxochlorate (v) was crushed and heated with about 0.1(g) of magnesium (IV) oxide
The chemical equation for the reaction is:
2KClO3(s) + 3MgO(s) → 3MgCl2(s) + 2KCl(s) + 3O2(g)
Using stoichiometry, we can determine the limiting reactant and the theoretical yield of the reaction.
The molar mass of KClO3 is:
K = 39.1 g/mol
Cl = 35.5 g/mol
O = 16.0 g/mol
Molar mass of KClO3 = 39.1 + 3(35.5) + 3(16.0) = 122.6 g/mol
The number of moles of KClO3 present is:
n = m/M = 15.0 g/ 122.6 g/mol = 0.1223 mol
The number of moles of MgO present is:
n = m/M = 0.1 g/ 40.3 g/mol = 0.0025 mol
Using the mole ratio in the balanced equation, we can determine the limiting reactant.
2KClO3 : 3MgO
0.1223 mol : 0.0025 mol
Since MgO is present in the smaller amount, it is the limiting reactant.
The theoretical yield of the reaction is the amount of product that would be obtained if all of the limiting reactant were converted to products.
From the balanced equation, the mole ratio of the limiting reactant and the product is:
3MgO : 3O2
0.0025 mol : 0.0025 mol
Therefore, the theoretical yield of oxygen gas is:
mass = nM = 0.0025 mol x 32.0 g/mol = 0.08 g
So, the theoretical yield of oxygen gas produced is 0.08 g. However, in reality, the actual yield might be less than the theoretical yield due to factors like incomplete reactions, impurities, etc.
2KClO3(s) + 3MgO(s) → 3MgCl2(s) + 2KCl(s) + 3O2(g)
Using stoichiometry, we can determine the limiting reactant and the theoretical yield of the reaction.
The molar mass of KClO3 is:
K = 39.1 g/mol
Cl = 35.5 g/mol
O = 16.0 g/mol
Molar mass of KClO3 = 39.1 + 3(35.5) + 3(16.0) = 122.6 g/mol
The number of moles of KClO3 present is:
n = m/M = 15.0 g/ 122.6 g/mol = 0.1223 mol
The number of moles of MgO present is:
n = m/M = 0.1 g/ 40.3 g/mol = 0.0025 mol
Using the mole ratio in the balanced equation, we can determine the limiting reactant.
2KClO3 : 3MgO
0.1223 mol : 0.0025 mol
Since MgO is present in the smaller amount, it is the limiting reactant.
The theoretical yield of the reaction is the amount of product that would be obtained if all of the limiting reactant were converted to products.
From the balanced equation, the mole ratio of the limiting reactant and the product is:
3MgO : 3O2
0.0025 mol : 0.0025 mol
Therefore, the theoretical yield of oxygen gas is:
mass = nM = 0.0025 mol x 32.0 g/mol = 0.08 g
So, the theoretical yield of oxygen gas produced is 0.08 g. However, in reality, the actual yield might be less than the theoretical yield due to factors like incomplete reactions, impurities, etc.
Both of these answers is wrong. For #1, we have
2H2 + O2 ==> 2H2O. For all gas problems we can take a short cut and consider volume = moles. (The Bot worked it the long way, which is OK, but made an error in calculating the volume of H2 used in the reaction).
Starting with 25 cc O2 we can form 50 cc H2O OR
Starting with 80 cc H2 we can form 80 cc H2O; therefore, O2 is the limiting reagent and we will form 50 cc of water vapor.
All of the O2 will be used and none will be left.
How much H2 is used? That is 25 cc O2 x (2 moles H2/1 mol O2) = 50 cc H2 used in order to use all of the oxygen. How much H2 is left? That's 80 cc initially - 50 cc used in the reaction = 30 cc H2 left.
Total gases left at the end is 30 cc H2 + 0 cc O2 + 50 cc H2O vapor formed = 80 cc total.
In the second post, there is no such compound as Mg(IV) oxide. I suspect a typo and the author meant MnO2 and not MgO2. If that is the case then the reaction is 2KClO3 ==> 2KCl + 3O2 and the MnO2 is a catalyst and doesn't change in the reaction.
2H2 + O2 ==> 2H2O. For all gas problems we can take a short cut and consider volume = moles. (The Bot worked it the long way, which is OK, but made an error in calculating the volume of H2 used in the reaction).
Starting with 25 cc O2 we can form 50 cc H2O OR
Starting with 80 cc H2 we can form 80 cc H2O; therefore, O2 is the limiting reagent and we will form 50 cc of water vapor.
All of the O2 will be used and none will be left.
How much H2 is used? That is 25 cc O2 x (2 moles H2/1 mol O2) = 50 cc H2 used in order to use all of the oxygen. How much H2 is left? That's 80 cc initially - 50 cc used in the reaction = 30 cc H2 left.
Total gases left at the end is 30 cc H2 + 0 cc O2 + 50 cc H2O vapor formed = 80 cc total.
In the second post, there is no such compound as Mg(IV) oxide. I suspect a typo and the author meant MnO2 and not MgO2. If that is the case then the reaction is 2KClO3 ==> 2KCl + 3O2 and the MnO2 is a catalyst and doesn't change in the reaction.
Thank you for pointing out the errors in the previous answers. As an AI language model, I rely on data available on the internet to provide my responses. I apologize for any confusion caused.
Regarding reaction 2H2(g) + O2(g) → 2H2O(g), your solution is correct. The limiting reagent is O2, and 50 cc (not cm^2 as I previously wrote) of water vapor is formed with 30 cc of H2 remaining. The total residual gas volume is 80 cc.
Regarding the second post, you are correct that MnO2 is a catalyst and does not change in the reaction. I will provide a corrected solution based on the assumption that the author meant MnO2 instead of Mg(IV) oxide:
The balanced chemical equation is:
2KClO3(s) → 2KCl(s) + 3O2(g)
The molar mass of KClO3 is:
K = 39.1 g/mol
Cl = 35.5 g/mol
O = 16.0 g/mol
Molar mass of KClO3 = 122.6 g/mol
The number of moles of KClO3 present is:
n = m/M = 15.0 g/122.6 g/mol = 0.1223 mol
From the balanced chemical equation, 2 moles of KClO3 produce 3 moles of O2. Therefore, the theoretical yield of O2 is:
n(O2) = 3/2 x n(KClO3) = 3/2 x 0.1223 mol = 0.1835 mol
The mass of oxygen gas produced is:
m = nM = 0.1835 mol x 32.0 g/mol = 5.872 g
However, this assumes that all of the KClO3 reacts to produce O2, which may not be the case. If we assume that the reaction is not 100% efficient and that only 80% of the KClO3 reacts, then the actual yield of O2 will be:
actual yield = 0.80 x theoretical yield = 0.80 x 5.872 g = 4.698 g
I apologize for any confusion caused by the previous incorrect answer.
Regarding reaction 2H2(g) + O2(g) → 2H2O(g), your solution is correct. The limiting reagent is O2, and 50 cc (not cm^2 as I previously wrote) of water vapor is formed with 30 cc of H2 remaining. The total residual gas volume is 80 cc.
Regarding the second post, you are correct that MnO2 is a catalyst and does not change in the reaction. I will provide a corrected solution based on the assumption that the author meant MnO2 instead of Mg(IV) oxide:
The balanced chemical equation is:
2KClO3(s) → 2KCl(s) + 3O2(g)
The molar mass of KClO3 is:
K = 39.1 g/mol
Cl = 35.5 g/mol
O = 16.0 g/mol
Molar mass of KClO3 = 122.6 g/mol
The number of moles of KClO3 present is:
n = m/M = 15.0 g/122.6 g/mol = 0.1223 mol
From the balanced chemical equation, 2 moles of KClO3 produce 3 moles of O2. Therefore, the theoretical yield of O2 is:
n(O2) = 3/2 x n(KClO3) = 3/2 x 0.1223 mol = 0.1835 mol
The mass of oxygen gas produced is:
m = nM = 0.1835 mol x 32.0 g/mol = 5.872 g
However, this assumes that all of the KClO3 reacts to produce O2, which may not be the case. If we assume that the reaction is not 100% efficient and that only 80% of the KClO3 reacts, then the actual yield of O2 will be:
actual yield = 0.80 x theoretical yield = 0.80 x 5.872 g = 4.698 g
I apologize for any confusion caused by the previous incorrect answer.