Moles 0_2_(g) React to Produce H_2O_(g)

[Eich]

How many moles of $H_{2}O_{(g)}$ are produced when 9.6 mol of $0_{2}_{(g)}$ react?

 

[The Bob]

How many moles of [tex]H_{2}O_{(g)}[/itex] are produced when 9.6 mol of $0_{2}_{(g)}$ react?

$$O_{2 (g)} \rightarrow H_2 O_{(g)}$$ is what you know so to balance (using x as unknowns) $$xH_{2(g)} + O_{2 (g)} \rightarrow xH_2 O_{(g)}$$
Now after the simple balancing apply what you know from the question so that (y being the value you should know): $$x_2 H_{2(g)} + y O_{2 (g)} \rightarrow x_2 H_2 O_{(g)}$$

I think that is the idea but someone might say otherwise.

Hope it helps.

The Bob (2004 ©)

EDIT: I removed the answer so that the original poster could try and solve it themselves. Sorry Gokul43201.

 

[danne89]

You may want to explicity say where the coefficent of the hydrogen came from.

 

[dextercioby]

And if i may ask,why would he do that?I think it's pretty obvious where & why that 19.2 came from...

Daniel.

 

[Eich]

Sorry sorry everyone. I know that was a stupid question. It's just the answer they gave was wrong.

Anyway it's $$9.6 mol O_{2} * 2 mol of H_{2}O over 1 mol of O_{2}$$
Which came to $$19 mol$$ with sig figs.

Sorry again. :/

 

[The Bob]

Which came to $$19 mol$$ with sig figs.

Why not have 19.2 moles?? It is more accurate. Significant figures will do very little to the answer.

The Bob (2004 ©)

 

[Eich]

Because our teacher is anal like that.

 

[The Bob]

Because our teacher is anal like that.

I see. I do hate teachers like that.

The Bob (2004 ©)

 

[Sirus]

The teacher is, nevertheless, correct. The value of 9.6 was given to two significant figures in the question, so the answer must be rounded to the same to represent the maximum certainty of the answer.