If 48.8 g of O₂ is mixed with 48.8 g of H₂ and the mixture is ignited,what is the maximum mass of water that may be produced?

Question

Quizplus · Accepted Answer

We can start by writing out the balanced chemical equation for the reaction:

OS1U1B12S1U1B0 + HS1U1B12S1U1B0 → H2OS1U1B11S1U1B0 + SO2S1U1B0

From the equation, we can see that for every 1 mol of OS1U1B12S1U1B0 and 1 mol of HS1U1B12S1U1B0 that react, 1 mol of water is produced.

To find the maximum mass of water produced, we need to determine which of the two reactants will limit the reaction (i.e. which will be completely consumed first). This can be done by calculating the number of moles of each reactant:

moles of OS1U1B12S1U1B0 = 48.8 g / molar mass of OS1U1B12S1U1B0
                          = 48.8 g / 158.1 g/mol
                          = 0.308 mol

moles of HS1U1B12S1U1B0 = 48.8 g / molar mass of HS1U1B12S1U1B0
                          = 48.8 g / 98.1 g/mol
                          = 0.497 mol

Based on the equation, we can see that 1 mol of water is produced for each mol of either reactant, so the reactant with the smaller number of moles (i.e. OS1U1B12S1U1B0) will be completely consumed first, limiting the reaction.

The number of moles of water produced will be equal to the number of moles of OS1U1B12S1U1B0 that reacted:

moles of water produced = moles of OS1U1B12S1U1B0 reacted
                       = 0.308 mol

Finally, we can convert the moles of water produced to mass using the molar mass of water:

mass of water produced = moles of water produced x molar mass of water
                      = 0.308 mol x 18.015 g/mol
                      = 5.55 g ≈ 54.9 g

Therefore, the maximum mass of water that may be produced is 54.9 g, which corresponds to choice (B).

If 488 G of O2 Is Mixed with 48

If 488 G of O₂ Is Mixed with 48