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

2 Answers
May 30, 2018

Approx. 89*g89g....

Explanation:

We got the stoichiometric equation...

H_2(g) + 1/2O_2(g) rarr H_2O(l)H2(g)+12O2(g)H2O(l)

"Moles of water"=(100*g)/(18.01*g*mol^-1)=5.55*molMoles of water=100g18.01gmol1=5.55mol

And, therefore, we need an equivalent quantity of dioxygen gas... a mass of 5.55*molxx1/2xx32.00*g*mol^-1=??*g5.55mol×12×32.00gmol1=??g..

May 30, 2018

177.62g (2.d.p)

Explanation:

Equation: 2H_2+O_2 -> H_2O2H2+O2H2O

First, we need to calculate the moles of water.

"moles" = "mass"/"molar mass"moles=massmolar mass
mass: 100g
molar mass: 1.008times2+16 = 18.0161.008×2+16=18.016

"moles"=100/18.016 = 5.55062167moles=10018.016=5.55062167

Now, we have to look at the ratio:

hydrogen:oxygen:water = 2:1:12:1:1

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 = 3216×2=32

Rearranging the equation gives me:

"mass" = "moles"times"molar mass"mass=moles×molar mass

"mass" = 5.55062167 times 32 = 177.62gmass=5.55062167×32=177.62g