No why are you asking this question?
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>
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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.
Answer is: density of the talc sample in air is 4,61 g/cm³.
m(talc) = 13,65 g.
d(talc) = 1,75 g/cm³.
V(talc) = m(talc) ÷ d(talc).
V(talc) = 13,65 g ÷ 1,75 g/cm³.
V(talc) = 7,8 cm³.
d(talc in air) = m(talc in air) ÷ V(talc).
d(talc in air) = 35,97 g ÷ 7,8 cm³.
d(talc in air) = 4,61 g/cm³.
The answer is dissolved salts
