You need to rethink this.
K = 273 + 34 which is NOT 315.4.
This is my last question and this one really threw me for a loop. I know that the temp has to be converted to kelvin which would make it 315.4
What volume of oxygen gas can be collected at 0.683 atm pressure and 34.0◦C when 42.4g of KClO3 decompose by heating, according to the following equation?
2KClO3(s)->2KCl(s) + 3O2(g)
Answer in units of L
2 answers
2KClO3(s)->2KCl(s) + 3O2(g)
mols KClO3 = grams molar mass = ?
Using the coefficients in the balanced equation, convert mols KClO3 to mols O2 gas. Then substitute n into PV = nRT and solve for V in liters at the conditions listed. Use P = 0.683 atm, R is 0.08206 L*atm/mol*K, T is 273 + 34 = ? and n from above.
mols KClO3 = grams molar mass = ?
Using the coefficients in the balanced equation, convert mols KClO3 to mols O2 gas. Then substitute n into PV = nRT and solve for V in liters at the conditions listed. Use P = 0.683 atm, R is 0.08206 L*atm/mol*K, T is 273 + 34 = ? and n from above.
1 answer