2H2 (g) + O2 (g)→ 2H2O (l)

A total of 12.0 grams of hydrogen and 64.0 grams of oxygen are available. Once mixed, this reaction runs to completion, meaning that all of the limiting reactant is consumed.
a. Which reactant is limiting?
b. How many moles of water are formed?
c. How many moles of hydrogen are consumed?
d. How many moles of oxygen are consumed? e. What is the mass of oxygen that is consumed?
f. What is the mass of water that is formed?

2H2 (g) + O2 (g)→ 2H2O (l)

A total of 12.0 grams of hydrogen and 64.0 grams of oxygen are available. Once mixed, this reaction runs to completion, meaning that all of the limiting reactant is consumed.
a. mols O2 = grams/molar mass = 64.0/32 = 2
mols H2 = 12.0/2 = 6
mols H2O formed from 2 mols O2 and excess H2 = 4
mols H2O formed from 6 mols H2 and excess O2 = 6
In limiting reagent problems (LR) the smaller number always wins because you never can get more then the smallest possible. So O2 is the LR.
b. 2 mols O2 x (2 mol H2O/1 mol O2) = 4 mols H2O formed.
c. 2 mol O2 x (2 mols H2/1 mol O2) = 4 mols H2 used.
d. all O2 is consumed; i.e., 2 mols.
e. all of it is 64.0 g
f. mass water = mols H2O x molar mass H2O = ?

a. To determine the limiting reactant, we need to compare the amounts of moles of hydrogen and oxygen.

1. Calculate the number of moles for each reactant:
Moles of hydrogen = mass of hydrogen / molar mass of hydrogen
Moles of oxygen = mass of oxygen / molar mass of oxygen

Given:
mass of hydrogen = 12.0 grams
molar mass of hydrogen (H2) = 2.02 g/mol
mass of oxygen = 64.0 grams
molar mass of oxygen (O2) = 32.00 g/mol

Moles of hydrogen = 12.0 g / 2.02 g/mol ≈ 5.94 mol
Moles of oxygen = 64.0 g / 32.00 g/mol = 2.00 mol

2. The limiting reactant is the reactant that produces the least amount of product. Since the ratio of hydrogen to oxygen in the balanced equation is 2:1, we can see that the moles of oxygen (2.00 mol) are less than half the moles of hydrogen (5.94 mol). Therefore, oxygen is the limiting reactant.

b. The balanced equation tells us that 2 moles of water are formed for every 2 moles of hydrogen consumed. Since the reaction goes to completion and all the limiting reactant is consumed, we can determine the number of moles of water formed using the moles of the limiting reactant (oxygen):

Moles of water formed = Moles of oxygen consumed = 2.00 mol (since the molar ratio is 2:1)

c. The moles of hydrogen consumed can be calculated using the balanced equation, which indicates a 2:2 stoichiometric ratio between hydrogen and water (2H2O):

Moles of hydrogen consumed = Moles of water formed = 2.00 mol

d. The moles of oxygen consumed can also be calculated using the balanced equation, which indicates a 1:1 stoichiometric ratio between oxygen and water:

Moles of oxygen consumed = Moles of water formed = 2.00 mol

e. To find the mass of oxygen consumed, we can use the moles of oxygen consumed:

Mass of oxygen consumed = Moles of oxygen consumed * molar mass of oxygen
Mass of oxygen consumed = 2.00 mol * 32.00 g/mol = 64.00 g

f. To find the mass of water formed, we can use the moles of water formed:

Mass of water formed = Moles of water formed * molar mass of water
Molar mass of water (H2O) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Mass of water formed = 2.00 mol * 18.02 g/mol = 36.04 g

To determine the limiting reactant and answer the subsequent questions, we need to follow these steps:

Step 1: Convert the given masses of hydrogen and oxygen into moles.

a. Hydrogen (H2):

12.0 grams of hydrogen × (1 mole/2.016 grams) = 5.95 moles of hydrogen

b. Oxygen (O2):

64.0 grams of oxygen × (1 mole/32.00 grams) = 2.00 moles of oxygen

Step 2: Examine the stoichiometric ratio of the balanced equation:

2H2 (g) + O2 (g) → 2H2O (l)

According to the equation, for every 2 moles of hydrogen consumed, 1 mole of oxygen is consumed, and 2 moles of water are formed.

Step 3: Calculate the amount of water that can be formed based on the limiting reactant.

a. Hydrogen:

Using the stoichiometric ratio from Step 2, the number of moles of water that can be formed from the given moles of hydrogen is:

5.95 moles of hydrogen × (2 moles of water/2 moles of hydrogen) = 5.95 moles of water

b. Oxygen:

Using the stoichiometric ratio from Step 2, the number of moles of water that can be formed from the given moles of oxygen is:

2.00 moles of oxygen × (2 moles of water/1 mole of oxygen) = 4.00 moles of water

Step 4: Compare the amounts of water that can be formed based on the limiting reactant.

The limiting reactant is the one that produces fewer moles of water when compared to the other reactant.

In this case, the number of moles of water formed from hydrogen is 5.95 moles, while the number of moles of water formed from oxygen is 4.00 moles.

Since oxygen produces fewer moles of water, it is the limiting reactant.

Now, we can answer the questions:

a. The limiting reactant is oxygen.

b. The number of moles of water formed is 4.00 moles.

c. The number of moles of hydrogen consumed is 2 times the number of moles of water formed, based on the stoichiometric ratio. Therefore, it is 2 × 4.00 moles = 8.00 moles.

d. The number of moles of oxygen consumed is the same as the number of moles of water formed since the stoichiometric ratio is 1:1. Therefore, it is 4.00 moles.

e. To calculate the mass of oxygen consumed, we can use its molar mass:

4.00 moles of oxygen × (32.00 grams/1 mole) = 128.00 grams of oxygen.

f. To calculate the mass of water formed, we can use its molar mass:

4.00 moles of water × (18.02 grams/1 mole) = 72.08 grams of water.