Unhealthy exhaust results from the combustion of octane (C_3H_10)in automobile e
ID: 1073470 • Letter: U
Question
Unhealthy exhaust results from the combustion of octane (C_3H_10)in automobile engines via the following reaction, Assuming the gasoline is pure octane and that a suburban dweller burn about 644 times 10^4 grams of octane (about 20 gallons) per week. How many grams of CO_2 are produced (per week) to warm the globe? At 25 degree C and 1.00 atm, how many ml of CO_2 are product? Give your anser in scientific notation. During combustion, Ne and O_2 in the air react to make the nasty brown gas NO_2 which is smog. The up into the atmosphere, combines with water vapor and plants. The following reactions are relevant. Balance reaction C Calculate the moles of HNO_3 that form if 22 times 10^4 moles of NO_2 rise up into the air and react completely with an unlimited amount of water vapor. A 200 mL sample of the lake water is collected titrated directly with 0.01M NaOH and it takes only 3 mL of the NAOH titrant to reach the equivalent point. Calculate the molarity (M) of HNO_3 in the lake water. Nitric acid is a strong acid and engages in the following proton transfer (dissociation) reaction with water (in which it is dissolved) suppose the concentration of HNO_3 in the lakewater is 0.08 mg/mL calculate the molarity of H_2O^- and the pH of the lakewater.Explanation / Answer
First calculate the moles of Octane
= amount in g / molar mass
= 6.44*10^4 g / 114 g/ mole
= 564.9 mole
Now calculate the moles of CO2 per week:
564.9 mole Octane *16 moleCO2 / 1 mole Octane
= 9038.6 mole CO2
Amount of CO2 per week
=9038.6 mole CO2 * molar mass;44 g/ mole
=3.98*10^5 g CO2
Now calculate the volume of CO2 as follows:
PV = n RT
Here T = 25 C = 298 K and P =1.0 atm
V = n RT / P
= 9038.6 mole CO2 *0.08206 L –atm / mole-K *298K/ 1.00 atm
= 2.21*10^5 L
= 2.21*10^8 ml
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