Answer to Question #221507 in General Chemistry for msa

Question #221507

Calculate the mass of water produced when 9.38 g

g of butane reacts with excess oxygen.


1
Expert's answer
2021-07-30T15:20:02-0400

"2C\n_4\nH\n_{10}\n(\ng\n)\n+\n13\n\nO\n_2\n(\ng\n)\n\u2192\n4\nC\nO\n_2\n(\ng\n)\n+\n5\nH\n_2\nO\n(\nl\n)"


Moles of butane "=\\frac{9.38}\n{58.12}=0.16 moles"


Moles of oxygen "=\\frac{13\u00d70.16}{2}=1.04moles"


Moles of water "=\\frac{5\u00d70.16}{2}=0.4moles"



Mass of water "=18.02\u00d70.4=7.208grams"


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