2. Given the following equation, C2H5OH(l) + 3 O2(g)---> 2 CO2(g) + 3 H2O(l)
How many liter of CO2 are expected in the combustion of 12.0 grams of ethanol?
My work:
12.0g C2H5OH X 1 mol C2H5OH/46.08g C2H5OH X 2 mol CO2/1 molC2H5OH X 22.4L CO2/1 mol CO2 = 11.7 L CO2
all i need to know is if the answre is correct
I didn't check the math but the procedure is right. Good work.
Your calculations appear to be correct. Let's go through it step by step to verify:
1. Begin with 12.0 grams of ethanol (C2H5OH).
2. Convert grams of ethanol to moles: 12.0 g C2H5OH * (1 mol C2H5OH / 46.08 g C2H5OH) = 0.26 mol C2H5OH.
3. Use the balanced equation to find the moles of CO2 produced: 0.26 mol C2H5OH * (2 mol CO2 / 1 mol C2H5OH) = 0.52 mol CO2.
4. Lastly, convert moles of CO2 to liters using the ideal gas law assumption: 0.52 mol CO2 * (22.4 L CO2 / 1 mol CO2) = 11.7 L CO2.
Therefore, your final answer is 11.7 L of CO2.