Consider the reaction
4 KO2(s) + 2 CO2(g) →
2 K2CO3(s) + 3 O2(g).
How much KO2 is needed to react with 75.0 L
of carbon dioxide at STP?
4KO2(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g).
mols CO2 = 75/22.4 = 3.34
mols KO2 needed = 3.34 mols CO2 x (4 mols KO2/2 mols CO2) =?
The grams KO2 = mols KO2 x molar mass KO2
Post your work if you get stuck.
Well, let's do some math! According to the balanced equation, we need 4 moles of KO2 to react with 2 moles of CO2. Since we know that 1 mole of any gas occupies 22.4 L at STP, we can calculate the moles of CO2 in 75.0 L by dividing 75.0 L by 22.4 L/mol.
Now, once we have the moles of CO2, we can use the stoichiometry of the equation to determine the moles of KO2 needed. For every 2 moles of CO2, we need 4 moles of KO2.
Finally, we convert moles of KO2 to grams by using its molar mass. Now take all this information, plug it into your calculator, and kindly ask your calculator to hold your hand while it does the math for you.
To find out how much KO2 is needed to react with 75.0 L of carbon dioxide at STP, we need to use the ideal gas law equation PV = nRT.
First, let's determine the number of moles of CO2 present in 75.0 L at STP.
STP stands for Standard Temperature and Pressure, which is defined as 0 degrees Celsius (273.15 K) and 1 atmosphere (1 atm).
Using the ideal gas law equation PV = nRT, we can rearrange the equation to solve for n (the number of moles):
n = PV / RT
where:
P = pressure (1 atm)
V = volume (75.0 L)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (273.15 K)
Plugging in the values:
n = (1 atm) * (75.0 L) / (0.0821 L·atm/mol·K * 273.15 K)
n = 2.930 mol
Now, let's use the balanced chemical equation to determine the stoichiometric ratio between KO2 and CO2:
4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)
From the equation, we can see that it takes 2 moles of KO2 to react with 1 mole of CO2.
Therefore, to find out the amount of KO2 needed to react with 2.930 mol of CO2, we can use the ratio:
2.930 mol CO2 * (2 mol KO2 / 1 mol CO2) = 5.860 mol KO2
So, 5.860 mol of KO2 is needed to react with 75.0 L of carbon dioxide at STP.
To determine how much KO2 is needed to react with 75.0 L of carbon dioxide at STP, we need to use the stoichiometry of the balanced reaction equation.
First, let's identify the coefficients in the balanced equation:
4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)
According to the balanced equation, 4 moles of KO2 react with 2 moles of CO2. This gives us a mole ratio of 4:2.
To calculate the amount of KO2 needed, we can set up a proportion using the mole ratio:
4 moles KO2 / 2 moles CO2 = x moles KO2 / 75.0 L CO2
Simplifying the proportion:
(4/2) moles KO2 = x moles KO2 / 75.0 L CO2
Now, let's solve for x moles KO2:
(4/2) moles KO2 = x moles / 75.0 L CO2
Cross-multiplying and solving for x:
(4/2) * 75.0 L CO2 = x moles KO2
150 L CO2 = x moles KO2
Therefore, to react with 75.0 L of carbon dioxide at STP, you would need 150 moles of KO2.