Question

Which of the following is the correct "setup" for the problem "How many grams of H2O will be produced from 2.1 moles of O2 and an excess of H2S?" according to the reaction 2H2S + 3O2 --> 2H2O + 2SO2.
A.  This is the same answer written slightly differently: 2.1 moles O2 x (18.02 g water/ 2 moles water)

B. The same answer written slightly differently: 2.1 moles O2 x (32.00 g O2/ 1 mole O2) x (18.02g water/ 32.00 g O2)
C. The same answer written slightly differently: 2.1 moles O2 x (2 moles water/ 3 moles O2) x (18.02 g water/ 1 mole water)
D. This is the same answer written a little bit differently: 2.1 moles O2 x (32.00 g/ 1 mole O2) x ( 2 moles water/ 3moles O2)

Answer 1

C is the correct answer.

You can see from the reaction that for every 3 moles of O2, 2 moles of water are formed.

So, for 2.1 moles of O2, 2.1 moles of O2 X (2 moles of water/3 moles of O2) will be formed.

Now, molar mass of H2O is 18.02 g.

So, multiply the moles of water by 18.02 g and you get your result.

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