How many moles of water are produced when 1.05 g of oxygen gas reacts with 1.22 g of hydrogen gas to produce water?

1.05 g of O2 is 1.05/32 = 0.0328125 moles

1.22 g of H2 is 1.22/2.016 = 0.60516 moles

O2 is the limiting reactant. There is more than enough H2 to combine with the available oxygen. Each mole of O2 forms 2 moles of H2O, according to the reaction

2H2 = O2 -> 2H2O

Therefore, double 0.0328125 moles
to get the answer

To find the number of moles of water produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed, determining the maximum amount of product that can be formed.

We can start by calculating the number of moles for each reactant:

1. Oxygen gas (O₂):
moles = mass / molar mass
moles = 1.05 g / (32.00 g/mol)
moles = 0.0328 mol

2. Hydrogen gas (H₂):
moles = mass / molar mass
moles = 1.22 g / (2.02 g/mol)
moles = 0.604 mol

To determine the limiting reactant, we need to compare the ratios of the coefficients in the balanced chemical equation. The balanced equation for the reaction is:

2H₂ + O₂ → 2H₂O

The ratio of moles of O₂ to H₂ required for the reaction is 1:2. So, for every 1 mole of O₂, we need 2 moles of H₂. Let's compare:

- Oxygen gas (O₂): 0.0328 mol
- Hydrogen gas (H₂): 0.604 mol / 2 = 0.302 mol

Since we have more moles of hydrogen gas than oxygen gas, the oxygen gas is the limiting reactant.

According to the balanced equation, 2 moles of water are produced for every 1 mole of O₂. Therefore, the number of moles of water produced can be determined as follows:

moles of water = moles of O₂ × (2 moles of H₂O / 1 mole of O₂)
moles of water = 0.0328 mol × 2
moles of water = 0.0656 mol

Therefore, 0.0656 moles of water are produced when 1.05 g of oxygen gas reacts with 1.22 g of hydrogen gas to produce water.

To find the number of moles of water produced, you need to first determine the balanced chemical equation for the reaction between oxygen gas (O2) and hydrogen gas (H2) to produce water (H2O).

The balanced chemical equation is:

2H2 + O2 → 2H2O

From the balanced equation, we can see that 2 moles of water are produced for every 1 mole of oxygen gas.

Now, let's calculate the moles of oxygen gas and hydrogen gas given in the question:

Moles of oxygen gas (O2):
Given mass of oxygen gas = 1.05 g
Molar mass of oxygen gas (O2) = 32.00 g/mol

Number of moles of oxygen gas = mass / molar mass
= 1.05 g / 32.00 g/mol

Moles of hydrogen gas (H2):
Given mass of hydrogen gas = 1.22 g
Molar mass of hydrogen gas (H2) = 2.02 g/mol

Number of moles of hydrogen gas = mass / molar mass
= 1.22 g / 2.02 g/mol

Since the balanced equation shows that oxygen gas reacts with hydrogen gas in a 1:2 ratio, you need to compare the moles of oxygen gas and hydrogen gas to determine the limiting reagent.

The limiting reagent is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

In this case, you have 1.05 g of oxygen gas and 1.22 g of hydrogen gas. To determine which one is the limiting reagent, compare the moles of each reactant:

Moles of oxygen gas = 1.05 g / 32.00 g/mol
Moles of hydrogen gas = 1.22 g / 2.02 g/mol

Compare the moles of oxygen gas with the moles of hydrogen gas. Whichever reactant has the smaller number of moles is the limiting reagent.

Now that you have determined the limiting reagent, you can calculate the moles of water produced. Since the balanced equation tells us that 2 moles of water are produced for every 1 mole of oxygen gas, we can use the mole ratio to find the moles of water.

Moles of water = Moles of limiting reagent (in this case, oxygen gas) * (2 moles of water / 1 mole of oxygen gas)

Insert the values into the equation to find the moles of water produced.