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Calculate the mass of water produced when 7.49 g of butane reacts with excess oxygen.

User Smedegaard
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1 Answer

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22 votes

Answer:


m_(H_2O)=12.9gH_2O

Step-by-step explanation:

Hey there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the reaction whereby butane is combusted in the presence of excess oxygen:


2C_4H_(10)+13O_2\rightarrow 8CO_2+10H_2O

Thus, we can evidence a 2:10 mole ratio of butane to water, and thus, the stoichiometric setup to calculate the mass of produced water is:


m_(H_2O)=7.49gC_4H_(10)*(1molC_4H_(10))/(52.12gC_4H_(10)) *(10molH_2O)/(2molC_4H_(10))*(18.02gH_2O)/(1molH_2O)\\\\m_(H_2O)=12.9gH_2O

Regards!

User Johnlinp
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