In the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor. Hydrogen + Oxygen #-># water. If you burn 58.1 g of hydrogen and produce 519 g of water how much oxygen reacted?

Answer 1

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

Clearly, approx. #500# #g# of dioxygen reacted.

The stoichiometrically balanced equation tells us unequivocally that #1# #mol# dihydrogen reacts with #0.50# #mol# dioxygen to give #1# #mol# water.
#"Moles of dihydrogen"# #=# #(58.1*g)/(2.02*g*mol)# #=# #28.7*mol# #H_2#
#"Moles of water"# #=# #(519.0*g)/(18.02*g*mol)# #=# #28.7*mol# #H_2O#
The water is stoichiometric with respect to the amount of dihydrogen. (What do I mean by this?) There was thus a stoichiometric quantity of dioxygen gas, i.e. #14.4*mol# #O_2#
#"Mass of dioxygen"# #=# #14.4*cancel(mol)xx32.00*g*cancel(mol^-1)# #=# #??g#
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Answer 2

29.05 g of oxygen reacted.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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