# Mass of H2 in a balloon

### Question

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

• Why are you doubling the mass of H2 in the final step of your calculation? Where does the factor (x2) come from? – Arsak May 15 at 4:27
• 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. – Arsak May 15 at 4:37
• 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. – andselisk May 15 at 4:39
• Thank you for adding the units.That was my bad. – Nathanael Calinisan Miranda May 15 at 4:50
• The answer sheet is wrong. – Chet Miller May 15 at 12:43