Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
Answer:
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Explanation:
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Answer:
Major organic products are- (a) propan-1-ol and (b) 2-methylpropan-2-ol
Explanation:
methyl magnesium bromide gives nucleophilic addition reaction with carbonyl group. Because methyl magnesium bromide is a strong nucleophile and carbonyl group is a strong electrophilic center.
Propanal contains an aldehyde group and propanone contains a ketone group. hence they both give nucleophilic addition with methyl magnesium bromide.
Dilute acid is added to protonate the alkoxide produced during nucleophilic addition.
Reactions are shown below.
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.
