A balloon of diameter $\pu{2 m}$ has a volume of $\pu{4190 dm^3}.$ It floats since it is given an up thrust equal to the mass of air it displaces.

(Given: average $M_\mathrm{r}$ of air = 29. Under conditions of inflation, $\pu{1 mol}$ of $\ce{H2}$ occupies a volume of $\pu{23 dm^3}).$

Find mass of $\ce{H2}$ in balloon.

My method

$$\text{Amount of}~\ce{H2} = \frac{\pu{4190 dm^3}}{\pu{23 dm3 mol-1}} = \pu{182.2 mol}$$ $$\text{Mass of}~\ce{H2}~\text{in balloon} = \pu{182.2 mol} × \pu{2 g mol-1} = \pu{364.4 g}$$

Correct answer: $\pu{182.2 g}$.

I don't understand why the answer is 182.2g

  • $\begingroup$ Why are you doubling the mass of H2 in the final step of your calculation? Where does the factor (x2) come from? $\endgroup$ – Arsak May 15 '19 at 4:27
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    $\begingroup$ Sure, I just didn't want to guess your thoughts from the numbers (because the guess could be wrong). It might help to edit the post and add a short explanation and the units. $\endgroup$ – Arsak May 15 '19 at 4:37
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    $\begingroup$ At first glance your solution looks reasonable. I allowed myself to add missing units – please don't drop them and always carry values and units together. $\endgroup$ – andselisk May 15 '19 at 4:39
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    $\begingroup$ Thank you for adding the units.That was my bad. $\endgroup$ – Nathanael Calinisan Miranda May 15 '19 at 4:50
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    $\begingroup$ The answer sheet is wrong. $\endgroup$ – Chet Miller May 15 '19 at 12:43

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