2. Carbon monooxide CO is product of incomplete combustion of fuel. Find the volume that 42g of CO gas occupies at STP

well, one moles occupies 22.4 dm^3 at STP.

How many moles is 42g?

1 mole of a gas (28 grams for CO) occupies 22.4 L at STP.

So do I do 22.4x42? Because when i did this is got 940.8. So is this my answer?

To find the volume that 42g of CO gas occupies at STP (Standard Temperature and Pressure), we can use the Ideal Gas Law equation:

PV = nRT

Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

At STP, the conditions are:
- Temperature (T) = 273.15 K
- Pressure (P) = 1 atmosphere
- R = 0.0821 L·atm/(mol·K)

First, we need to convert the mass of CO gas (42g) to moles.

The molar mass of CO (carbon monoxide) is:
Molar mass(C) + Molar mass(O) = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol

Now, we can calculate the number of moles (n) using the following equation:
n = mass / molar mass

n = 42g / 28.01 g/mol = 1.498 moles (rounded to 3 decimal places)

Now that we know the number of moles of CO gas, we can substitute these values into the Ideal Gas Law equation:

PV = nRT

V = (nRT) / P

V = (1.498 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm

V ≈ 34.99 liters (rounded to 2 decimal places)

Therefore, 42g of CO gas occupies approximately 34.99 liters at STP.