Answer:
Explanation:
<u>1. Chemical quation</u>
The reaction of aluminium, sodium hydroxide and water is represented by the balanced chemical equation:
- 2Al(s) + 2NaOH(s) + 6H₂O(l) → 2Na[Al(OH)₄] (aq) + 3H₂(g) ↑
The coefficients of each reactant and product give the theoretical mole ratios.
To find the limiting reactant you compare the theoretical ratios with the ratio of the available substaces.
<u>2. Theoretical mole ratio:</u>
- 2 mol Al : 2 mol NaOH : 6 mol H₂O
Equivalent to
- 1 mol Al : 1 mol NaOH : 3 mol H₂O
<u>3. Actual ratio</u>
a) Convert each mass to number of moles
Formula:
- number of moles = mass in grams / molar mass
Al:
- molar mass = atomic mass = 26.982g/mol
- number of moles = 51.0g / 26.982g/mol = 1.89 mol
NaOH:
- number of moles = 84.1g / 39.997g/mol = 2.10 mol
H₂O:
- number of moles = 25.0g / 18.015g/mol = 1.39 mol
Divide all the mole amounts by the least number:
- Al: 1.89/1.39 = 1.36
- NaOH: 2.10 = 1.52
- H₂O: 1.39 = 1.00
- 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O
<u>4. Comparison</u>
<u />
Theoretical ratio:
- 1 mol Al : 1 mol NaOH : 3 mol H₂O
Actual ratio:
- 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O
Multiply by 3:
- 4.08 mol Al : 4.56 mol NaOH : 3.00 mol H₂O
Now, yo can see that the first two are in excess with respect the third one, making that the water consumes first, before any of the other two consumes. Therefore, the limiting reactant is the water.
Easy (:
C₃H₈ (g) + O₂ (g) ---> CO₂ (g) + H₂O
1 C₃H₈ (g) + 5 O₂ (g) ---> 3 CO₂ (g) + 4 H₂O
With this there is the same amount of elements on each side (You take the larger number in front and multiply it by whats after it. So you have 3 C, 8 H, and 10 O on one side, and then the next mirrors is with 3 C, 8 H, and 10 O.
Carbon monoxide<span>, lead, </span>nitrogen dioxide<span>, </span>ozone<span>, particulate matter, and </span>sulfur dioxide
Answer:
256 g
Step-by-step explanation:
a) Balanced equation
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 2.016 32.00 18.02
2H₂ + O₂ ⟶ 2H₂O
m/g: 32.3 288
(b) Moles of H₂O
n = 288 g H₂O × (1 mol H₂O/18.02 g H₂O)
= 15.98 mol H₂O
(c) Moles of O₂
The molar ratio is (2 mol H₂O /1 mol O₂)
n = 15.98 mol H₂O × (1 mol O₂ /2 mol H₂O)
= 7.991 mol O₂
(d) Mass of O₂
m = 7.991 mol O₂× (32.00 g O₂/1 mol O₂)
= 256 g O₂
The reaction requires 256 g O₂.
<em>Note</em>: If you base your answer on the mass of <em>hydrogen</em>, you get the same result.
Mass of O₂ = 32.3 g H₂ × 1 mol H₂/2.016 g H₂
= 16.02 mol H₂ × 1 mol O₂ /2 mol H₂
= 8.011 mol O₂ × 32.00 g O₂/1 mol O₂
= 256 g O₂
