2CO(g) + O2(g) ==> 2CO2(g)
A. If 16.01 L of oxygen is reacted at STP, how many moles of carbon dioxide is produced?
B. If 1.722 mol of carbon monoxide is reacted at STP, what volume of carbon dioxide is produced?
C. At STP, what volume of oxygen is needed to produce 82.1 g of carbon dioxide?;
D. What volume of carbon monoxide at STP is needed to produce 663 mL of carbon dioxide?
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To answer these questions, we first need to determine the stoichiometry of the balanced chemical equation:
2CO(g) + O2(g) → 2CO2(g)
A. To find the number of moles of carbon dioxide produced when 16.01 L of oxygen is reacted at STP, we can use the Ideal Gas Law. At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L.
The molar ratio between oxygen and carbon dioxide is 1:2. From the balanced equation, we can see that for every two moles of CO2 produced, one mole of O2 is consumed. Therefore, we can set up a proportion:
(16.01 L O2 / 22.4 L O2) = (x moles CO2 / 2 moles CO2)
Solving for x, we find:
x = (16.01 L O2 / 22.4 L O2) * 2 moles CO2
x ≈ 1.43 moles CO2
B. To find the volume of carbon dioxide produced when 1.722 mol of carbon monoxide is reacted at STP, we can use the same stoichiometry calculations. The molar ratio between carbon monoxide and carbon dioxide is 2:2 (or 1:1).
So, we can set up the following proportion:
(1.722 mol CO / 1 mol CO) = (x L CO2 / 2 L CO2)
Solving for x, we find:
x = (1.722 mol CO / 1 mol CO) * 2 L CO2
x = 3.444 L CO2
C. To find the volume of oxygen needed to produce 82.1 g of carbon dioxide at STP, we need to convert the mass of CO2 to moles and then use the stoichiometry of the balanced equation to find the volume of oxygen.
First, we calculate the number of moles of CO2:
82.1 g CO2 * (1 mol CO2 / 44.01 g CO2) = 1.866 mol CO2
Since the molar ratio between CO2 and O2 is 2:1 (two moles of CO2 per one mole of O2), we can set up the following proportion:
(1.866 mol CO2 / 2 mol CO2) = (x L O2 / 1 L O2)
Solving for x, we find:
x = (1.866 mol CO2 / 2 mol CO2) * 1 L O2
x ≈ 0.933 L O2
D. To find the volume of carbon monoxide needed to produce 663 mL of carbon dioxide at STP, we need to convert the volume of CO2 to moles and then use the stoichiometry of the balanced equation to find the volume of carbon monoxide.
First, we convert the volume of CO2 from mL to L:
663 mL CO2 * (1 L / 1000 mL) = 0.663 L CO2
Next, we calculate the number of moles of CO2:
0.663 L CO2 * (1 mol CO2 / 22.4 L CO2) = 0.0296 mol CO2
Since the molar ratio between CO and CO2 is 1:2 (or 0.5:1), we can set up the following proportion:
(0.0296 mol CO2 / 2 mol CO2) = (x L CO / 0.5 L CO)
Solving for x, we find:
x = (0.0296 mol CO2 / 2 mol CO2) * 0.5 L CO
x = 0.0074 L CO