To get the percent yield, we will use this formula:
((Actual Yield)/(Theoretical Yield)) * 100%
Values given: actual yield is 220.0 g
theoretical yield is 275.6 g
Now, let us substitute the values given.
(220.0 grams)/(275.6 grams) = 0.7983
Then, to get the percentage, multiply the quotient by 100.
0.7983 (100) = 79.83%
Among the choices, the most plausible answer is 79.8%
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Answer:
330 mL of (NH₄)₂SO₄ are needed
Explanation:
First of all, we determine the reaction:
(NH₄)₂SO₄ + 2NaOH → 2NH₃ + 2H₂O + Na₂SO₄
We determine the moles of base:
(First, we convert the volume from mL to L) → 62.6 mL . 1L/1000 mL = 0.0626L
Molarity . volume (L) = 2.31 mol/L . 0.0626 L = 0.144 moles
Ratio is 2:1. Therefore we make a rule of three:
2 moles of hydroxide react with 1 mol of sulfate
Then, 0.144 moles of NaOH must react with (0.144 .1) /2 = 0.072 moles
If we want to determine the volume → Moles / Molarity
0.072 mol / 0.218 mol/L = 0.330 L
We convert from L to mL → 0.330L . 1000 mL/1L = 330 mL
<u>Given:</u>
Mass of calcium nitrate (Ca(NO3)2) = 96.1 g
<u>To determine:</u>
Theoretical yield of calcium phosphate, Ca3(PO4)2
<u>Explanation:</u>
Balanced Chemical reaction-
3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2
Based on the reaction stoichiometry:
3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2
Now,
Given mass of Ca(NO3)2 = 96.1 g
Molar mass of Ca(NO3)2 = 164 g/mol
# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles
Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles
Molar mass of Ca3(PO4)2 = 310 g/mol
Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g
Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g
Isotopes of elements have same atomic numbers and different mass number(atomic masses).
Correct answer is option E. <span>It is a redox reaction in which Zn is oxidized at the anode, and V is reduced at the cathode.
Reason:
In above reaction, the oxidation state of VO3- is +5, while that of VO2 is +4. Thus there is reduction of V from +5 to +4
In case of Zn, oxidation state of Zn is increased from 0 to +2, Thus process is referred as oxidation. </span>
