how many grams of iron are needed to react with 31.0L of chlorine gas at STP to produce iron(III) chloride?
Convert L Cl2 gas to moles remembering that there are 22.4 L/mol at STP. Then follow the directions in this example.
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To determine the number of grams of iron needed to react with chlorine gas and produce iron(III) chloride, you'll need to use the balanced chemical equation for the reaction and apply the concept of stoichiometry.
The balanced equation for the reaction is:
2 Fe(s) + 3 Cl2(g) → 2 FeCl3(s)
From the equation, we can see that 2 moles of iron react with 3 moles of chlorine gas to produce 2 moles of iron(III) chloride.
To find the number of moles of chlorine gas, we can use the ideal gas law at STP (standard temperature and pressure). At STP, 1 mole of any ideal gas occupies 22.4 liters.
31.0 liters of chlorine gas at STP:
Moles of Cl2 = volume / molar volume
Moles of Cl2 = 31.0 L / 22.4 L/mol
Moles of Cl2 ≈ 1.384 moles
Using the stoichiometry of the balanced equation, we can determine the moles of iron needed to react with the moles of chlorine gas:
2 moles of Fe : 3 moles of Cl2 :: x moles of Fe : 1.384 moles of Cl2
By cross-multiplying, we can find:
x = (2 moles of Fe * 1.384 moles of Cl2) / 3 moles of Cl2
x ≈ 0.9227 moles of Fe
Finally, to calculate the grams of iron required, we can use the molar mass of iron (Fe):
Grams of Fe = moles of Fe * molar mass of Fe
Grams of Fe = 0.9227 moles * 55.85 g/mol
Grams of Fe ≈ 51.47 grams
Therefore, approximately 51.47 grams of iron are needed to react with 31.0L of chlorine gas at STP to produce iron(III) chloride.
Note: It's always a good practice to round the final answer to an appropriate number of significant figures based on the given data.
To determine how many grams of iron are needed to react with chlorine gas, we first need to write a balanced equation for the reaction. The balanced equation for the reaction between iron and chlorine gas to produce iron(III) chloride is:
2 Fe(s) + 3 Cl2(g) -> 2 FeCl3(s)
From the balanced equation, we can see that the molar ratio between iron and chlorine gas is 2:3. This means that for every 2 moles of iron, we need 3 moles of chlorine gas.
To find the amount of chlorine gas in moles, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (at STP, pressure is 1 atm)
V = volume (31.0 L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (at STP, temperature is 273 K)
Rearranging the equation to solve for n, we have:
n = PV / RT
Substituting the given values, we get:
n = (1 atm) * (31.0 L) / (0.0821 L·atm/mol·K * 273 K)
n ≈ 1.208 moles
Since the molar ratio between iron and chlorine gas is 2:3, we can set up a proportion to find the moles of iron needed:
2 moles Fe / 3 moles Cl2 = x moles Fe / 1.208 moles Cl2
Cross-multiplying and solving for x, we get:
x = (2 moles Fe * 1.208 moles Cl2) / 3 moles Cl2
x ≈ 0.802 moles Fe
Finally, to calculate the grams of iron needed, we need to use the molar mass of iron, which is approximately 55.85 g/mol:
grams of iron = moles of iron * molar mass of iron
= 0.802 moles * 55.85 g/mol
≈ 44.79 grams
Therefore, approximately 44.79 grams of iron are needed to react with 31.0 L of chlorine gas at STP to produce iron(III) chloride.