Chemical Equation Lab Report

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The chemical equation provides a variety of quantitative information for the calculations of combining weights of materials involved in a chemical process. The coefficients of a balanced equation tell us the relative formula units of reactants and products participating in a reaction. For instance:

C_8 H_18+12½O_2→8〖CO〗_2+9H_2 O … (3.7)

Note:
(i) A balanced equation remains valid if each of the coefficients in Eq. (3.7) is multiplied by the same number.
(ii) The basis selected for calculations is mass rather than moles.
(a) If mass is given, the given mass is converted to moles using molecular weight.
(b) Then using appropriate stoichiometric ratio, obtain required moles of products and reactants.
(c) Change moles of
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Calculate the following items: (a) The percentage of excess carbon furnished, based on the principal reaction. (b) The percentage conversion of Fe2O3 to Fe. (c) The pounds of carbon used up and the pounds of CO produced per ton of Fe2O3 charged. (d) What is the selectivity in this process (of Fe with respect to FeO)?
Solution
Setting up the mole equivalent for the principal reaction as follows; Fe_2 O_3 + 3C → 2Fe + 3CO
Initial mass: 1 100 lb 300 lb - -
Mass at time, t: - - 600 lb
Molar mass: 160 12 56
Number of moles: 6.875 25.00 10.714
Mole equivalent: 6.875 8.333 5.357 In addition, setting up the mole equivalent for the undesired side reaction as follows; Fe_2 O_3 + C → 2FeO + CO
Initial mass: 1 100 lb 300 lb - -
Mass at time, t: - - 91.5 lb
Molar mass: 160 12 72
Number of moles: 6.875 25.00
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91.5 lb/72 (molar mass of FeO)) ∴Selectivity of Fe=(10.714 moles Fe)/(1.271 moles FeO)=8.43 (mole Fe)⁄(mole FeO)

Example 3.6: Chemical Equation and Stoichiometry
Question
A common method used in manufacturing sodium hypochlorite bleach is by the reaction: Cl_2+2NaOH→NaCl+NaOCl+H_2 O
Chlorine gas is bubbled through an aqueous solution of sodium hydroxide, after which the desired product is separated from the sodium chloride (a by-product of the reaction). An aqueous solution of NaOH containing 520.45 kg of pure NaOH is reacted with 386.82 kg of gaseous chlorine to give 280.91 kg of NaOCl. (a) What was the limiting reactant? (b) What was the percentage excess of the excess reactant used? (c) What is the degree of completion of the reaction, expressed as the moles of NaOCl formed to the moles NaOCl that would have formed if the reaction had gone to completion? (d) What is the yield of NaOCl per amount of chlorine used (on a weight basis)?
Solution
First, setting up the mole equivalent as follows:
Cl_2 + 2NaOH→NaCl+NaOCl+H_2 O
Initial mass (kg): 386.82 520.45 - -
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