Answer to Question #170067 in General Chemistry for Lari

Question #170067

Please with the steps..


Consider the mixing of 5.1 L of  CO2 (g) and 3.1 L of H2O (g) at 45°C and 850 mm Hg. Determine the mass of O2 (g) that can be produced from the unbalanced reaction:


CO2 (g) + H2O (g) = C4H10 (l) +O2 (g)


Mass= ? in grams





1
Expert's answer
2021-03-10T06:16:36-0500

To determine the mass of oxygen, the number of moles of gaseous "CO_2" and "H_2O" have to be computed based on the ideal gas law:

"pV=nRT"


"n=\\frac{pV}{RT}\\\\\nn(CO_2)=\\frac{850\\times 133 \\times 10^{-3} Pa \\times 5.1 L}{8.314 J \\cdot K^{-1} \\cdot mol^{-1} * (45+273) K} = 0.22 mol \\\\\nn(H_2O)=\\frac{850\\times 133 \\times 10^{-3} Pa \\times 3.1 L}{8.314 J \\cdot K^{-1} \\cdot mol^{-1} * (45+273) K} = 0.13 mol"


If we want to know the mass of oxygen that can be produced from the reaction, it has to be balanced first:


"8CO_2 (g) + 10H_2O (g) =2C_4H_{10} (l) +13O_2 (g)"


The limiting reagent is water:

"CO_2 : 0.22\/8 = 0.027\\\\\nH_2O: 0.13 \/ 10 = 0.013"


Now, the mass of oxygen is:

"m(O_2) = n (O_2)\\times M (O_2)= \\frac{13\\times 0.13}{10} mol \\times 32 = 5.41 g"



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