Identify the oxidizing agent and the reducing agent for Ca(s)+Fe2+(aq)→Ca2+(aq)+Fe(s)
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Answer:
Explanation:
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In this case, since the combustion of the hamburger released 335.4 kcal of energy and that energy is received by the calorimeter, we can write:
And the heat of the calorimeter is written in terms of the temperature change and the calorimeter constant:
Thus, given the released heat by the hamburger due to its combustion and the temperature change, Cv for the calorimeter turns out:
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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.