What is the mass of water vapor produced when 3.2 liters reacts with 8.7 liters of oxygen gas at STP?

2.57g

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2H2 + O2 —> 2H2O

Next let us determine the limiting reactant. This is achieved as follows:

From the equation,

2L H2 required 1L of O2.

Therefore, 3.2L of H will require = 3.2/2 = 1.6L of O2

From the calculation above, O2 is excess because the volume of O2 given from the question is far greater than the volume of O2 obtained from our calculation. Therefore, H2 is the limiting reactant.

Now let us covert 3.2L of H2 to mole. This is illustrated below:

1mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2 will occupy 3.2L i.e

Xmol of H2 = 3.2/22.4 = 0.143mol

From the equation,

2moles of H2 produced 2moles of H2O.

Therefore, 0.143mol of H2 will also produce 0.143moles of H2O.

Now, we can obtain the mass of the water vapour produced by convert 0.143mol of H2O to gram. This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Number of mole of H2O = 0.143mol

Mass of H2O =?

Mass = mole x Molar Mass

Mass of H2O = 0.143 x 18 = 2.57g

The mass of water vapour produce is 2.57g