[Shipwreck]

A hydrogen-filled balloon was ignited and 1,50 grams of hydrogen reacted with 12 g of oxygen. How many grams of water vapor were formed?

The answer in the back of the book is 13,50 g, but I get a slightly different answer (13,40 grams):

The reaction is 2H₂ + O₂  2H₂O

1,50 g H₂ / 2,016 g/mol (H₂) = ,744047619 mol H₂

12 g O₂ / 31,998 g(mol (O₂) = ,375 mol O₂

Hydrogen is therefore the limiting reagent since the molar ratios are 2:1 hydrogen to oxygen. (,375 x 2 = ,75...there is an excess of oxygen)

,744047619 / 2 = ,372. Only ,372 mol of oxygen is consumed during the reaction even though there are ,375 moles present:

,372 mol O₂ x 31,998 g/mol = 11,9 g O₂

11,9g + 1,5g (since all the hydrogen gas is consumed in the reaction) = 13,4g of water.

Could someone please give me a hint as to where I made a mistake? Thanks in advance!

[AWK]

Frankly saying, both answers are not exactly correct. You did calculation correctly, but since mass of oxygen shows 2 significant digits, absolutely correct answer should be rounded to 13. Your answer is correct for mass 12.0.
In textbook, rounded masses (but treated as exact numbers) of H₂ and O₂  were used (2 and 32, respectively), and stricly correct answer should be rounded to 14.

[Shipwreck]

In textbook, rounded masses (but treated as exact numbers) of H₂ and O₂  were used (2 and 32, respectively), and stricly correct answer should be rounded to 14.

Hello and thanks for your reply. I'm curious as to why the textbook would round the molar masses to 2 and 32? Also I realized that I made a mistake in copying the question: the amount of oxygen is actually 12,00g so my answer is actually 13,40g.

[AWK]

1.50 g of H₂ suggests then 3 significant digits.
Rounding some masses to whole numbers is sufficient to obtain result with error significantly below 5 %. Even for mass of Cl=35 the error is below 2%.