Question

Question #13dca

Answer 1

20 grams

Explanation:

`H_2+1/2 O_2→H_2O`

molecular weights
`H_2` = 2 g / mol
`O_2` = 32 g / mol
`H_2O` = 18 g / mol

• First, you need to find the limiting reactant. We can do this by calculating how many grams of `H_2O` would be produced per reactant.

Hydrogen:

4 g `H_2` * `(1 mol H_2) / (2 g H_2)` * `(1 mol H_2O) / (1 mol H_2)` * `(18 g H_2O) / (1 mol H_2O)` = 36 g `H_2O`

Oxygen:

20 g `O_2` * `(1 mol O_2) / (36 g O_2)` * `(1 mol H_2O) / (0.5 mol O_2)` * `(18 g H_2O) / (1 mol H_2O)` = 20 g `H_2O`

• From this, we can see that oxygen is the limiting reactant. Therefore, the amount of `H_2O` produced is 20 grams.

Answer 2

`22.52` g (2.d.p.)

Explanation:

First we need to find the limiting reactant. We know the water molecule contains twice as many hydrogen atoms as it does oxygen.

Number of moles of hydrogen we have:

Number of moles X molar mass = mass in grams:
So number of moles = mass in grams / molar mass

So we have 4g of hydrogen:

moles = `4/1.008= 3.96825` (5.d.p.)

Number of moles of oxygen is:

moles = `20/15.999= 1.25008`

In order for a full reaction we would need twice as many moles of hydrogen as there is oxygen.

`2 xx 1.25008=2.5006` moles.

We have more than this, so oxygen is the limiting reactant.

To find the total amount of product we can make, we calculate the required ratio:

`(O)/(H_2) = 1/2= 1.25008/x=> x=2 xx1.25008= 2.50016` moles

Mass of oxygen = 20g

Mass of hydrogen = `2.50016 xx 1.008 = 2.52016` g

Mass of `H_2O = 20 + 2.52016 = 22.52016`g