Question

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

Answer 1

[AI-generated answer — Bot]

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

Answer 2

15.0(g)of potassium trioxochlorate (v) was crushed and heated with about 0.1(g) of magnesium (IV) oxide

Answer 3

[AI-generated answer — Bot]

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.

Answer 4

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.

Answer 5

[AI-generated answer — Bot]

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.

Answer 6

[AI-generated answer — Step-by-Step Bot]

To determine the volume of residual gases, we need to consider the stoichiometry of the reaction between hydrogen (H₂) and oxygen (O₂).

From the balanced equation:
2H₂(g) + O₂(g) → 2H₂O(g)

We can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

First, let's calculate the number of moles of hydrogen and oxygen used in the reaction.

Given:
Volume of hydrogen gas (H₂) = 80 cm²
Volume of oxygen gas (O₂) = 25 cm³

To convert the given volumes to the number of moles, we will use the ideal gas law equation:

PV = nRT

Where:
P = pressure (assumed constant)
V = volume
n = number of moles
R = ideal gas constant
T = temperature (assumed constant)

Since we have constant pressure and temperature, we can rewrite the equation as:

V/n = constant

Therefore, the ratio of the volumes to the number of moles is constant.

Using this ratio, we can calculate the number of moles of hydrogen and oxygen:

Number of moles of hydrogen (nH₂):
nH₂ = V1 / VmH₂

Where:
V1 = volume of hydrogen gas
VmH₂ = molar volume of hydrogen gas at standard temperature and pressure (STP = 22.4 L/mol)

Substituting the values:
V1 = 80 cm² = 0.08 L
VmH₂ = 22.4 L/mol

nH₂ = 0.08 L / 22.4 L/mol = 0.00357 mol

Number of moles of oxygen (nO₂):
nO₂ = V2 / VmO₂

Where:
V2 = volume of oxygen gas
VmO₂ = molar volume of oxygen gas at STP (22.4 L/mol)

Substituting the values:
V2 = 25 cm³ = 0.025 L
VmO₂ = 22.4 L/mol

nO₂ = 0.025 L / 22.4 L/mol = 0.00112 mol

Since the stoichiometric ratio is 2:1 for H₂:O₂, we need twice the number of moles of oxygen to react completely with the moles of hydrogen.

Therefore, the number of moles of oxygen required for the reaction is 2 * nH₂ = 2 * 0.00357 mol = 0.00714 mol.

Since we only have 0.00112 mol of oxygen, there will be an excess of oxygen left after the reaction.

To calculate the volume of residual gases, we need to find the difference between the moles of oxygen used in the reaction and the moles of oxygen initially present:

Residual moles of oxygen = moles of oxygen initially present - moles of oxygen used in the reaction
Residual moles of oxygen = 0.00112 mol - 0.00714 mol = -0.00602 mol

Since the calculated value is negative, it means that there is an excess of oxygen. However, the value doesn't make physical sense since we can't have a negative amount of molecules.

Therefore, we can conclude that there won't be any residual gases left after the reaction since the reaction is stoichiometrically balanced.

Answer 7

[AI-generated answer — Explain Bot]

To calculate the volume of residual gases, we need to determine which gas is in excess and which gas is limiting. The limiting reactant is the one that is completely consumed, while the excess reactant is left over.

Given that 80 cm^2 of hydrogen (H2) gas and 25 cm^3 of oxygen (O2) gas were reacted to form water (H2O) gas, we need to compare the amount of each gas used.

To do this, we need to convert the provided values into a common unit. The unit of volume is already consistent (cm^3), so no conversion is needed.

The balanced chemical equation for the reaction is:
2H2(g) + O2(g) → 2H2O(g)

From the equation, we can see that for every 2 moles of hydrogen gas, we need 1 mole of oxygen gas to react completely.

1 mole of any ideal gas at standard temperature and pressure (STP) occupies 22.4 L or 22,400 cm^3. This is known as molar volume.

First, let's calculate the number of moles for each reactant:
Moles of hydrogen = 80 cm^3 / 22,400 cm^3/mol = 0.00357 mol
Moles of oxygen = 25 cm^3 / 22,400 cm^3/mol = 0.00112 mol

Now, let's compare the moles of hydrogen and oxygen to determine the limiting reactant.

Since the reaction ratio is 2:1 (hydrogen to oxygen), we divide the moles of oxygen by 0.00112 mol to find the ratio:
0.00357 mol / 0.00112 mol = 3.183

This ratio tells us that there are approximately 3 moles of hydrogen for every mole of oxygen. Since the actual ratio is 2:1, we can see that oxygen is the limiting reactant.

Now that we know oxygen is the limiting reactant, we need to calculate the amount of excess hydrogen remaining after the reaction.

Since 2 moles of hydrogen are required to react with 1 mole of oxygen, and we have determined that oxygen is the limiting reactant, we can infer that only half the amount of hydrogen is required.

Therefore, the number of moles of hydrogen consumed = 0.5 * 0.00112 mol = 0.00056 mol.

Now, let's calculate the volume of residual gases.

Moles of hydrogen remaining = Moles of hydrogen initially - Moles of hydrogen consumed
Moles of hydrogen remaining = 0.00357 mol - 0.00056 mol = 0.00301 mol

Now, we convert moles to volume using the molar volume at STP:

Volume of hydrogen remaining = Moles of hydrogen remaining * 22,400 cm^3/mol
Volume of hydrogen remaining = 0.00301 mol * 22,400 cm^3/mol = 67.43 cm^3

Therefore, the volume of residual gases (hydrogen) is approximately 67.43 cm^3.