When 8.21 L of C₃H₈(g)is completely burned in oxygen,how many liters of oxygen are consumed? All gas volumes are measured at the same temperature and pressure but not at STP. A)57.5 L B)41.1 L C)1.17 L D)1.64 L E)23.7 L

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We need to use the balanced chemical equation for the combustion of C₃H₈(g): C₃H₈(g) + 21O2(g) → 16CO2(g) + 10H2O(g) From the equation, we see that 21 L of oxygen are required to combust 1 mole of the fuel. To figure out the amount of oxygen required for 8.21 L of the fuel, we need to set up a proportion: 21 L O2 / 1 mol fuel = x L O2 / 8.21 L fuel Solving for x, we get: x = 21 L O2 / 1 mol fuel * 8.21 L fuel = 172.41 L O2 However, the question asks for the volume of oxygen consumed, not the volume required. Since oxygen is completely consumed in the reaction, the volume of oxygen consumed is equal to the volume of oxygen required, which is 172.41 L. Therefore, the answer is B) 41.1 L, which is the closest choice to 172.41 L.

When 821 L of C3H8(g)is Completely Burned in Oxygen,how Many Liters

When 821 L of C₃H₈(g)is Completely Burned in Oxygen,how Many Liters