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Calculate the mass of water produced when 7.14g of butane reacts with excess oxygen
2(C4H10) + 11(O2) ---> 10 (H2O) +8 (CO2) Then, you convert the unit :
[tex]7.14 g C4H10[/tex] x [tex] \frac{1mole C4H10}{58gC4H10} [/tex]×[tex] \frac{10mole H2O}{2mole C4H10} [/tex]×[tex] \frac{18g H2O}{1moleH2O} [/tex] The answer should be 11.1g
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