Molar mass for (NH4)2SO4
1 answer:
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The theoretical yield of will be 4.20 grams while the percent yield will be 7.93%
From the equation of the reaction, the mole ratio of to
is 1:2.
Mole of 48.1 mL, 0.478 M = 0.478 x 48.1/100 = 0.023 mols
Mole of 48.8 mL, 0.160 M = 0.160 x 48.8/1000 = 0.0078 moles
is the limiting reactant.
Mole ratio of and
= 1:1
Equivalent mole of = 0.023 moles
Mass of 0.023 noles = 0.023 x 182.81 = 4.20 grams
With 0.0333 g of recovered, percent yield = 0.333/4.2 x 100 = 7.93%
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Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.
SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)
Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.
1) Calculate the theoretical yield of SnS2. ;
2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.
Answer: Option (A) is the correct answer.
Explanation:
According to the Gibb's free energy conditions:
If < 0, then reaction is spontaneous.
If > 0, then reaction is non-spontaneous.
If = 0, then reaction is equilibrium.
Since, the given reaction is . It denotes that liquid state of fluorine is changing into gaseous state. And, the boiling point of fluorine is very low.
Therefore, at low and high temperature fluorine is a gas. Hence, the reaction is spontaneous in nature.
Thus, we can conclude that the statement at low temperature, the reaction is spontaneous and < 0 and at high temperature, the reaction is spontaneous and
< 0, is correct.