Answer:
[tex]m_{H_2O}=6.84gH_2O[/tex]
Explanation:
Hello.
In this case, for 5.17 g of hexane (molar mass = 86 g/mol) and 16.5 g of oxygen (molar mass =32 g/mol), we have the reaction:
[tex]C_6H_{14}+\frac{19}{2} O_2\rightarrow 6CO_2+7H_2O[/tex]
Next, via the 1:7 mole ratio of hexane to carbon dioxide and 19/2:7 mole ratio of oxygen to carbon dioxide, we compute the yielded mass of water (molar mass = 18 g/mol) as its theoretical yield by the two masses of reactants and we infer that the limiting reactant is that yielding the fewest moles of product:
[tex]m_{H_2O}^{by\ Hexane}=5.17gC_6H_{12}*\frac{1molC_6H_{12}}{86 gC_6H_{12}} *\frac{7molH_2O}{1molC_6H_{12}}*\frac{18gH_2O}{1molH_2O} =7.57gH_2O\\\\m_{H_2O}^{by\ Oxygen}=16.5gO_2*\frac{1molO_2}{32gO_2} *\frac{7molH_2O}{\frac{19}{2}molO_2 }*\frac{18gH_2O}{1molH_2O} =6.84gH_2O[/tex]
Whereas it is evidenced that oxygen yields the fewest grams of water, therefore, it is the limiting reactant and the theoretical yield of water is:
[tex]m_{H_2O}=6.84gH_2O[/tex]
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