Answer:
[tex]m_{CO_2}=3.709gCO_2 \\\\m_{H_2O}=1.898gH_2O[/tex]
Explanation:
Hello.
In this case, since the molecular formula of the given alcohol is C₄H₁₀O (molar mass = 74.14 g/mol), we can write its combustion reaction as shown below:
[tex]C_4H_1_0O+6O_2\rightarrow 4CO_2+5H_2O[/tex]
Thus, since there is a 1:4 mole ratio with carbon dioxide (molar mass = 44..01 g/mol) and a 1:5 mole ratio with water (molar mass = 18.02 g/mol), we can compute the obtained masses as shown below:
[tex]m_{CO_2}=1.562gC_4H_1_0O*\frac{1mol}{74.14gC_4H_1_0O} *\frac{4molCO_2}{1molC_4H_1_0O} *\frac{44.01gCO_2}{1molCO_2}=3.709gCO_2 \\\\m_{H_2O}=1.562gC_4H_1_0O*\frac{1mol}{74.14gC_4H_1_0O} *\frac{5molH_2O}{1molC_4H_1_0O} *\frac{18.02gH_2O}{1molH_2O}=1.898gH_2O[/tex]
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