Answer:
65.4%
Explanation:
The redox reaction is a 1:1:1 reaction because the reagents suffer a double displacement reaction, and the substance that is substituted have the same charge (H+ and Br-), thus, we first need to know which of the reagents is the limiting.
Let's test the 4-nitrobenzaldehyde as the limiting. The mass needed for sodium borohydride (m) is the mass given of 4-nitrobenzaldehyde multiplied by the stoichiometric mass of sodium borohydride divided by the stoichiometric mass of 4-nitrobenzaldehyde. The stoichiometric mass is the number of moles in the stoichiometric representation (1:1:1) multiplied by the molar mass, so:
m = (4.13 * 37.83*1)/(151.12*1)
m = 1.034 g
So, the mass needed of the other reagent is larger than the mass that was given, so, it will be the limiting, and the stoichiometric calculus must be done with it.
The mass of the product that was expected is then:
m = (0.700*153.14*1)/(37.83*1)
m = 2.83 g
The percent yield is the mass that was formed divided by the expected mass, and then multiplied by 100%:
%yield = (1.85/2.83)*100%
%yield = 65.4%
Answer:
21.6 g
Explanation:
The reaction that takes place is:
First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
- 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂
0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.6 mol CH₄ *
= 1.2 mol H₂O
Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:
- 1.2 mol * 18 g/mol = 21.6 g
Answer:
B. A precipitate will form since Q > Ksp for calcium oxalate
Explanation:
Ksp of CaC₂O₄ is:
CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻
Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:
Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹
In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.
Replacing in Ksp formula:
[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.
If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.
Thus, right answer is:
<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>
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