Robert S. answered • 04/10/21

Patient PhD Chemist with 40 Years of R&D And Teaching

Hello, Lisa,

Let's start by balancing the equation. We can see that, as written, we have one oxygen coming in, but two (O_{2}) in the product side.

H_{2}O → H_{2 }+ O_{2}

Let's try making it 2 H_{2}O and see what happens:

2H_{2}O → H_{2 }+ O_{2}

That took care of the O's, but now we have two extra H's coming in. So let's add a "2" to the H_{2} to the product side:

2H_{2}O → 2H_{2 }+ O_{2}

Now it is balanced. 4 hydrogens and 2 oxygens coming in, and 4 hydrogens and 2 oxygens going out.

Now we can calculate the amounts of both products. We are given 10.0 grams of H_{2}O (note that we have 3 sig figs here). Convert that to moles H_{2}O by dividing it by water's molar mass of 18 grams/mole. That gives us 0.556 moles of water. We can see from the balanced equation that we only get 1 mole of O_{2} for every 2 moles of H_{2}O, a molar ratio of 1/2. So if we start with 0.556 moles of water, we'll get (1/2)*(0.556 moles) of O_{2}. That is 0.278 moles of O_{2}. Convert that to grams by multiplying it by oxygen's molar mass of 18 grams/mole: 8.89 grams O_{2}.

Do the same type of analysis for H_{2}. Here, the molar ratio of H_{2}/H_{2}O is 1 = we get the same number of moles of H_{2} as we have of H_{2}O coming in, which for us is 0.556 moles. Convert that to grams hydrogen by multiplying by hydrogen's molar mass of 2 grams/mole. That gives us 1.11 grams of H_{2}.

A nifty check on our work would be to see if the mass of the reactants equals the mass of the products, as it should due to the conservation of mass law.

O_{2} 8.89 grams

H_{2} 1.11

H_{2})L 10.0 grams

It works!

Bob