Question

1.562 g sample of the alcohol CH3CHOHCH2CH3 is burned in an excess of oxygen. What masses of H2O and CO2 should be obtained

Answer 1

`m_(CO_2)=3.709gCO_2 \\\\m_(H_2O)=1.898gH_2O`

Step-by-step 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:

`C_4H_1_0O+6O_2\rightarrow 4CO_2+5H_2O`

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:

`m_(CO_2)=1.562gC_4H_1_0O*(1mol)/(74.14gC_4H_1_0O) *(4molCO_2)/(1molC_4H_1_0O) *(44.01gCO_2)/(1molCO_2)=3.709gCO_2 \\\\m_(H_2O)=1.562gC_4H_1_0O*(1mol)/(74.14gC_4H_1_0O) *(5molH_2O)/(1molC_4H_1_0O) *(18.02gH_2O)/(1molH_2O)=1.898gH_2O`

Best regards!