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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%
More on yields of reactions can be found here: brainly.com/question/17042787
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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.
Explanation:
one thing to know is that higher surface area = higher boiling point.
NaCl has the smallest surface area, so it's the first one.
H2O has less surface area than methane, so it's second.
Methane has more surface area than H20, so it's third.
The big molecule has the most surface area, so it's last
