The balanced equation is Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂.
Next step is to convert everything to moles.
12.6g Fe₂O₃ x (1 mol Fe₂O₃ / 159.7g Fe₂O₃) = 0.07890 mol Fe₂O₃
9.65g CO x (1 mol CO / 28.01g CO) = 0.3445 mol CO
The third step is to determine the limiting and excess reactants.
0.07890 mol Fe₂O₃ x (3 mol CO/1 mol Fe₂O₃) = 0.2367 mol CO
Therefore Fe₂O₃ is the limiting reagent while CO is in excess.
0.07890 mol Fe x (2 mol Fe(s) / 1 mol Fe₂O₃) = 0.1578 mol Fe(s)
0.1578 mol Fe x (55.84g Fe / mole Fe) = 8.812g Fe is the theoretical yield
%yield = (7.23g / 8.812g) x 100% = 82.0% is the percent yield
<h2>
Answer: 6 moles</h2>
<h3>
Explanation:</h3>
3 H₂ + N₂ → 2 NH₃
↓ ↓
4 mol 3 mol
Since the moles of N₂ is the smaller of the two reactants, then N₂ is the limiting factor (the reactant that will decide how much ammonia is produced since it has the smaller amount of moles). ∴ we have to use it in calculating the number of moles of ammonia
The mole ratio of N₂ to NH₃ based on the balanced equation is 1 to 2.
∴ the moles of NH₃ = moles of N₂ × 2
= 3 moles × 2
= 6 moles
Answer: I am confident the answer is B
Explanation:
forgive me if im wrong
Answer:
The new temperature is 373 K
Explanation:
Step 1: Data given
Volume air = 5000 mL = 5.0 L
Temperature = 223K
New volume = 8.36 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒V1 = the initial volume = 5.0 L
⇒T1 = the initial temperature = 223 K
⇒V2 = the new volume = 8.36 L
⇒T2 = the new temperature
5.0/223 = 8.36 /T2
T2 = 373 K
The new temperature is 373 K
Answer:
There are 1.05 x 10²⁴ molecules in 48.6 g N₂
Explanation:
1 mol of N₂ has a mass of (14 g * 2) 28 g.
Then, 48.6 g of N₂ will be equal to (48.6 g *(1 mol/ 28 g)) 1.74 mol.
Since there are 6.022 x 10²³ molecules in 1 mol N₂, there will be
(1.74 mol *( 6.022 x 10²³ / 1 mol)) 1.05 x 10²⁴ molecules in 1.74 mol N₂ (or 48. 6 g N₂).
