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

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

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 :)

To answer the first question about whether the balloon will float or not, we need to compare the density of the balloon filled with CO gas to the density of air.

The density of air at 25.0 °C and 1.00 atm is given as 1.17 g/L.

To find the density of the CO gas-filled balloon, we need to calculate the mass of the gas.

Given that the empty balloon weighs 10.0 g and is filled to a volume of 20.0 L, we can assume that the weight of the gas is negligible compared to the weight of the balloon. So, we can ignore the weight of the balloon itself and consider only the weight of the gas.

The total mass of the CO gas-filled balloon is then given as 10.0 g (mass of the empty balloon).

Now, we can find the density of the CO gas-filled balloon by dividing the mass by the volume:

Density of CO gas-filled balloon = 10.0 g / 20.0 L = 0.50 g/L

Comparing this density to the density of air (1.17 g/L), we can see that the density of the CO gas-filled balloon is lower. Therefore, the balloon will float in air.

Regarding the second question about determining the volume of gas produced from the complete combustion of 1.0 kg of methane (CH4) at 20.0 °C and 1.00 atm.

To solve this, we need to use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to find the number of moles of methane.

Given that the molar mass of methane is 16.04 g/mol, we can calculate the number of moles by dividing the mass by the molar mass:

Number of moles of methane = 1.0 kg / 16.04 g/mol = 62.19 mol

Now, we can substitute the values into the ideal gas law equation:

(1.00 atm) * V = (62.19 mol) * (0.0821 L·atm/mol·K) * (293.15 K)

Simplifying the equation, we obtain:

V = (62.19 mol * 0.0821 L·atm/mol·K * 293.15 K) / 1.00 atm

V = 1.54 x 10^3 L

Therefore, the correct answer is approximately 1.54 x 10^3 L, not 1.50 x 10^3 L as stated in your answer.