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

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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:

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

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