Using the equation 2 H2 + O2 ---> 2 H2O determine how many grams of oxygen will be needed to form 100.0 grams of water?

Question

16821 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-30T06:56:50

Approx. $89*g$ ....

We got the stoichiometric equation...

$H_2(g) + 1/2O_2(g) rarr H_2O(l)$

$"Moles of water"=(100*g)/(18.01*g*mol^-1)=5.55*mol$

And, therefore, we need an equivalent quantity of dioxygen gas... a mass of $5.55*molxx1/2xx32.00*g*mol^-1=??*g$ ..

Answer 2 · 2018-05-30T07:00:07

177.62g (2.d.p)

Equation: $2H_2+O_2 -> H_2O$

First, we need to calculate the moles of water.

$"moles" = "mass"/"molar mass"$
mass: 100g
molar mass: $1.008times2+16 = 18.016$

$"moles"=100/18.016 = 5.55062167$

Now, we have to look at the ratio:

hydrogen:oxygen:water = $2:1:1$

Therefore, looking at water and oxygen, they have the same ratio of 1:1. That means that the moles of water is equal to the moles of oxygen.

Using the same equation as earlier, we already know the following:
moles = 5.55062167
molar mass = $16times2 = 32$

Rearranging the equation gives me:

$"mass" = "moles"times"molar mass"$

$"mass" = 5.55062167 times 32 = 177.62g$