Consider the balanced equation for the following reaction:
3Ca(ClO3)2(aq) + 2Li3PO4(aq) → Ca3(PO4)2(s) + 6LiClO3(aq)
Determine the theoretical yield of LiClO3(aq) in grams if the percent yield of LiClO3(aq) is 81.0% and 9.45 moles of LiClO3(aq) forms.

1055.83g

Explanation:

Data obtained from the question include:

Percentage yield of LiClO3 = 81.0%

Actual yield of LiClO3 = 9.45 moles

Now, let us convert 9.45 moles of LiClO3 to grams. This is illustrated below:

Molar Mass of LiClO3 = 7 + 35.5 + (16x3) = 7 + 35.5 + 48 = 90.5g/mol

Number of mole of LiClO3 = 9.45 moles

Mass of LiClO3 =?

Mass = number of mole x molar Mass

Mass of LiClO3 = 9.45 x 90.5

Mass of LiClO3 = 855.225g

Now we obtain the theoretical yield as follow:

%yield = Actual yield/Theoretical yield x 100

81% = 855.225/ Theoretical yield

81/100 = 855.225/ Theoretical yield

Cross multiply to express in linear form

Theoretical yield x 81= 100x855.225

Divide both side by 81

Theoretical yield = (100x855.225)/81

Theoretical yield = 1055.83g

Therefore, the theoretical yield of LiClO3 is 1055.83g

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a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.