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Answer 1 · 2017-08-02T11:29:34

#H_2 + 1/2 O_2 -> H_2O#

Atomic weights: 1 g (H) and 16 g (O).

You have excess oxygen since 16 grams of oxygen and 2 grams of hydrogen yield 18 grams of water. Therefore, your limiting reactant is oxygen.

If 16 grams of oxygen and 2 grams of hydrogen form 18 grams of #H_2O#

You wonder how much water is formed if you have 20 grams of oxygen and more than sufficient amount of hydrogen.

Your answer is

#=(20times18)/16 = 22.5# grams of water.

You will have an excessive amount of hydrogen which is 1.5 grams.

Answer 2 · 2017-08-02T11:31:20

Well, mass is conserved, so we are going to end up with #24*g# of oxygen and hydrogen, whatever the compound/elemental makeup.

We follow the stoichiometric equation......

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

ANd so dihydrogen, and dioxygen react in a 2:1 molar relationship....

#"Moles of dihydrogen"=(4*g)/(2.0*g*mol^-1)=2*mol#

#"Moles of dioxygen"=(20*g)/(32.0*g*mol^-1)=0.625*mol#

Clearly dioxygen is the limiting reagent, the reagent in stoichiometric deficiency, and thus, given the reaction we should form.......

#0.625*molxx2xx18.01*g*mol^-1=22.5*g# of water.

What is the mass of hydrogen left over?