What volume of oxygen at S•T•P would react with carbon to form 4•40g of CO2
According to the equation
C+O2>CO2
[O=16,C=12, 1 mole of a gas occupies 22•4dm3 at S•T•P
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.
To determine the volume of oxygen required, we need to calculate the moles of carbon dioxide produced first.
Step 1: Calculate the moles of CO2 produced:
Given that the mass of CO2 produced is 4.40g and the molar mass of CO2 is 44g/mol, we can use the equation:
moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 4.40g / 44g/mol
moles of CO2 = 0.10 mol (rounded to two decimal places)
Step 2: Determine the moles of oxygen required:
From the balanced equation, we can see that the stoichiometric ratio between carbon and oxygen is 1:1. Therefore, the moles of oxygen required will be the same as the moles of carbon dioxide produced.
moles of oxygen = 0.10 mol (from step 1)
Step 3: Convert moles of oxygen to volume at STP:
Given that 1 mole of a gas occupies 22.4 dm3 at STP, we can calculate the volume of oxygen required using the equation:
volume of oxygen = moles of oxygen × 22.4 dm3/mol
volume of oxygen = 0.10 mol × 22.4 dm3/mol
volume of oxygen = 2.24 dm3
Therefore, the volume of oxygen required to react with 4.40g of carbon to form 4.40g of CO2 at STP is 2.24 dm3.
To determine the volume of oxygen at STP that would react with carbon to form 4.40g of CO2, we need to calculate the number of moles of CO2 and then use the stoichiometry of the balanced equation to find the number of moles of oxygen. From there, we can calculate the volume using the given information that 1 mole of a gas occupies 22.4 dm3 at STP.
Let's start by finding the number of moles of CO2 formed from 4.40g.
Step 1: Calculate the molar mass of CO2:
C = 12.01 g/mol
O = 16.00 g/mol (2 oxygen atoms in CO2)
Molar mass of CO2 = (12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol
Step 2: Calculate the number of moles of CO2:
moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 4.40 g / 44.01 g/mol = 0.100 mol
Now that we have the number of moles of CO2, we can use the stoichiometry of the balanced equation to find the number of moles of oxygen.
The balanced equation is:
C + O2 -> CO2
From the equation, we can see that for every 1 mole of CO2 formed, 1 mole of oxygen reacts. Therefore, the number of moles of oxygen is equal to the number of moles of CO2.
moles of O2 = 0.100 mol
Finally, we can calculate the volume of oxygen at STP using the given information that 1 mole of a gas occupies 22.4 dm3 at STP.
volume of O2 = moles of O2 x 22.4 dm3/mol
volume of O2 = 0.100 mol x 22.4 dm3/mol = 2.24 dm3
Therefore, the volume of oxygen at STP that would react with carbon to form 4.40g of CO2 is 2.24 dm3.