Answer:
Mass = 357.7 g
Explanation:
Given data:
Mass of Fe = 250 g
Mass of oxygen = 120 g
Mass of iron(III) oxide produced = ?
Solution:
Chemical equation:
4Fe + 3O₂ → 2Fe₂O₃
Number of moles of Fe:
Number of moles = mass/molar mass
Number of moles = 250 g/ 55.8 g/mol
Number of moles = 4.48 mol
Number of moles of O₂ :
Number of moles = mass/molar mass
Number of moles = 120 g/ 32 g/mol
Number of moles = 3.75 mol
Now we will compare the moles of reactants with product.
Fe : Fe₂O₃
4 : 2
4.48 : 2/4×4.48 = 2.24
O₂ : Fe₂O₃
3 : 2
3.75 : 2/3×3.75= 2.5
Less number of moles of Fe₂O₃ are produced by Fe thus it will act as limiting reactant.
Mass of Fe₂O₃:
Mass = number of moles × molar mass
Mass = 2.24 mol × 159.69 g/mol
Mass = 357.7 g
Answer:
The answer to your question is 1.36 x 10²³ atoms
Explanation:
Data
number of atoms = ?
mass of the sample = 34.2 g
Molecule = Cl₂O₅
Process
1.- Calculate the molar mass of Cl₂O₅
Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g
2.- Calculate the atoms of Cl₂O₅
151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms
34.2 g of Cl₂O₅ ------------ x
x = (34.2 x 6.023 x 10²³) / 151
x = 1.36 x 10²³ atoms
Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s
