Fe + O2--->Fe2O3
How many grams of iron are required to convert 80 mL of oxygen at STP?
4Fe + 3O2 --> 2Fe2O3
mols O2 = 80 mL x (1 mol/22,400 mL) = ?
Using the coefficients in the balanced equation, convert mols O2 to mols Fe.
Then convert mols Fe to grams. g = mols x atomic mass Fe.
To calculate the grams of iron required to convert 80 mL of oxygen at STP, we need to use the balanced chemical equation and the molar ratios.
The balanced chemical equation is:
Fe + O2 ---> Fe2O3
From the equation, we can see that the ratio of Fe to O2 is 1:1. This means that for every 1 mole of Fe, we need 1 mole of O2.
To convert from mL to moles for O2, we need to use the Ideal Gas Law equation, which is:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume (80 mL)
n = number of moles (to be determined)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (STP = 273.15 K)
Rearranging the equation to solve for n, we have:
n = PV / RT
Substituting the values:
n = (1 atm) * (80 mL) / (0.0821 L·atm/mol·K * 273.15 K)
n = (80/1000 L) / (0.0821 L·atm/mol·K * 273.15 K)
n ≈ 0.0031 moles
Since the ratio of Fe to O2 is 1:1, we need 0.0031 moles of Fe to react with the given amount of O2.
To convert moles to grams, we need to use the molar mass of iron, which is 55.845 g/mol.
Thus, the mass of iron required is:
Mass of iron = 0.0031 moles * 55.845 g/mol
Mass of iron ≈ 0.173 g
Therefore, approximately 0.173 grams of iron are required to convert 80 mL of oxygen at STP.
To solve this question, we need to calculate the number of moles of oxygen and then use the balanced chemical equation to determine the number of moles of iron required. Finally, we can convert the moles of iron to grams.
Step 1: Calculate the number of moles of oxygen
To determine the number of moles, we'll use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the oxygen is at STP (Standard Temperature and Pressure), we can use the following values:
STP: T = 273.15 K, P = 1 atm, R = 0.0821 L·atm/(mol·K)
V = 80 mL = 80/1000 L (convert mL to L)
Now substitute the values into the equation:
(1 atm) (80/1000 L) = n (0.0821 L·atm/(mol·K)) (273.15 K)
Solving for n, we find:
n = (1 atm * 80/1000 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n ≈ 0.0035 moles of oxygen
Step 2: Use the balanced chemical equation to find the moles of iron
By examining the balanced chemical equation: Fe + O2 → Fe2O3, we can see that the stoichiometric ratio between iron and oxygen is 2:1. This means that 2 moles of iron react with 1 mole of oxygen.
Since we know that the oxygen is 0.0035 moles (determined in Step 1), the number of moles of iron needed will be half of that:
0.0035 moles / 2 = 0.00175 moles of iron
Step 3: Convert moles of iron to grams
To convert moles to grams, we need to know the molar mass of iron. Iron (Fe) has a molar mass of approximately 55.845 g/mol.
To calculate the mass of iron, we can use the formula:
Mass = Moles × Molar Mass
Mass = 0.00175 moles × 55.845 g/mol
Solving for Mass, we find:
Mass ≈ 0.097 g
Therefore, approximately 0.097 grams of iron are required to convert 80 mL of oxygen at STP.