Answer:
81.8 %
Explanation:
The balanced equation for the reaction is ⇒
So to find the number of moles of we do 6.04kg ÷ 2 = 3.02 moles.
The mole ratio between and is 3:2. So to find the number of moles you do × 2 which gives you 2.0124 moles.
Then you find the mass by doing moles multiplied by Mr: 2.0124 x 17 which gives you 34.21 kg.
Then to find the percentage yield you do: x 100 = 81.82.......
Which is 81.8 %
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Answers:
1.
The mole ratio of AlCl₃ to Cl₂ is 2 : 3. This means for production of every 2 moles of AlCl₃ 3 moles of Cl₂ is consumed. Hence, if 3 moles of AlCl₃ is produced then Cl₂ consumed is,
= 3 × 3 / 2
= 4.5 moles of Cl₂
2.
Mole ratio of C₃H₈ to CO₂ is 1 : 3.
Hence, 0.5 moles of C₃H₈ will produce 1.5 moles of CO₂.
Also,
At STP one mole of an ideal gas occupied 22.4 L of volume. Hence, 1.5 moles of CO₂ gas will occupy,
Volume = 1.5 mol × 22.4 L/mol
Volume = 33.6 L of CO₂
Answer:
17.136 g H2
Explanation:
First we need to convert moles of Zinc to moles of Hydrogen in order to find out how many moles of hydrogen are produced.
When we look at our balanced equation, we can see that for every mole of Zn reacting, there is 1 mol of H2 produced. (imagine a 1 in front of the elements that have no numbers in front of them)
8.5 mol Zn * = 8.5 mol H2
Now we will convert from moles of Hydrogen to grams
The atomic mass of H2 is 2(1.008) g/1 mol... so we will use this to convert to the number of grams of Hydrogen produced..
8.5 mol H2 * = 17.136 g H2
Explanation:
In a Brønsted-Lowry acid-base reaction, a conjugate acid is the species formed after the base accepts a proton. By contrast, a conjugate base is the species formed after an acid donates its proton.
Proton = H⁺
This means for the molecules that requires us to look for their conjugate bases, we simply remove a proton to it.
a. What are the conjugate bases of the molecules:
i C6H5OH : C6H5O⁻
ii. CH3-SH : CH3-S⁻
iii. CH3-CH2-CO2H : CH3-CH2-COO⁻
The molecules that requires us to look for their conjugate acids, we simply add a proton to it.
b. What are the conjugate acids of the molecules:
i. CH3–(CH2)-CO2- : i. CH3–(CH2)-COOH
ii. CH3–(CH2)-NH2 : ii. CH3–(CH2)-NH3⁺
iii. Ring at right ?