quick esstimate:
CO = 12+16 = 28 g/mol
about 22.4 liters/mol at stp
so we are talking around
28*20/22.4 = 25 grams of CO
+10 grams of rubber = 35 grams total
will the air lift that?
1.17*20 = 23.4 grams of buoyancy
no way, not close
1. An empty balloon weighs 10.0 g and is filled to a volume of 20.0 L with CO gas. If the density of air at 25.0 °C and 1.00 atm is 1.17 g/L, will the balloon float?
2. Determine the volume of gas produced at 20.0 °C and 1.00 atm from the complete combustion of 1.0 kg of methane (CH4).
My answer: 1.50 x 10^3 L
2 answers
CH4 = 12+4 = 16 g/mol
O2=32 g/mol
CO2 = 12+32 = 44 g/mol
CH4 + 2O2 ---> CO2 + 2H2O
I guess we want just the CO2 gas, although that H2O is initially steam
same mols of CO2 as mols of CH4
1000 grams / 16g/mol
= 1000/16 mols of CH4
so 1000/16 mols of CO2= 62.5 mols
quick estimate using 22.4 liters/mol
*22.4 liters/mol = 1400 liters
pretty close to your answer :)
O2=32 g/mol
CO2 = 12+32 = 44 g/mol
CH4 + 2O2 ---> CO2 + 2H2O
I guess we want just the CO2 gas, although that H2O is initially steam
same mols of CO2 as mols of CH4
1000 grams / 16g/mol
= 1000/16 mols of CH4
so 1000/16 mols of CO2= 62.5 mols
quick estimate using 22.4 liters/mol
*22.4 liters/mol = 1400 liters
pretty close to your answer :)
0 answers