Question #13dca

2 Answers
Oct 3, 2017

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.
Oct 3, 2017

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 = 22.52016g