What volume of oxygen at S•T•P would react with carbon to form 4•40g of CO2

Question

According to the equation
C+O2>CO2
[O=16,C=12, 1 mole of a gas occupies 22•4dm3 at S•T•P

Answer 1

To determine the volume of oxygen at STP that would react with 4.40g of CO2, we need to use stoichiometric calculations.

First, we need to calculate the number of moles of CO2 using its molar mass:

Molar mass of CO2 = (12g/mol * 1) + (16g/mol * 2) = 44g/mol

Number of moles of CO2 = 4.40g / 44g/mol = 0.1 mol

According to the balanced chemical equation, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of CO2.

Therefore, the molar ratio of oxygen to CO2 is 1:1.

Now, we can determine the volume of oxygen at STP using the ideal gas law:

PV = nRT

Since we want the volume in dm3, we can rewrite the equation as:

V = (n * R * T) / P

Plugging in the values:

V = (0.1 mol * 22.4 dm3/mol*K * 273 K) / 1 atm

V ≈ 6.03 dm3

Therefore, approximately 6.03 dm3 of oxygen at STP would react with 4.40g of CO2.