Item 9 Consider the reaction: AL+-Fe203 Also, + Fe A) Balance the equation. Writ
ID: 1009963 • Letter: I
Question
Item 9 Consider the reaction: AL+-Fe203 Also, + Fe A) Balance the equation. Write the coefficients from left to right, separated by commas. For example: 1,2,3 B) If you start with 62.5 g of Al and 133.0 g of Fe2Os, what is your limiting reactant? C) How many grams of Al O can you make? D) If you do the experiment in lab and produce 78.2 g of AlhO3, what is your percent yield? Item 10 Consider the molecules MX, and MX5. M has 5 valence electrons. X has 7 valence electrons. A) What is the molecular geometry (shape) of MXs? B) Is MX3 polar or non-polar? C) What is the molecular geometry (shape) of MX5? D) Is MXs polar or non-polar? Item 13 A) You need 650 mL of 2.70 M HCl solution for an experiment but the stock room only has 12.0 M HCl solution available. What volume of 12.0 M HCI do you need (in mL)? B) Explain how you would make the required 650 mL of 2.7 M HCl solution. Item 14 A) What is the molarity of a 575 mL solution that contains 42.7 g of NaCl? B) What mass of solid NaOH are needed to prepare 1.5 L of a 0.85 M solution?Explanation / Answer
For balancing a chemical equation, we just need to look at both the sides of the reaction and try to satisfy first the stoichiometry of one of them and accordingly we need to make the changes in the other and in the former as well.
The balanced chemical equation will be
2 Al + Fe2O3 ------> Al2O3 + 2 Fe
For deciding the limiting reagent, we need to evaluate the moles first and divide it with its stoichiometry from the balanced equation . After doing this, the one with the less value will be the limiting reagent.
no .of moles of Al = 62.5 / 26.98 = 2.3165 moles ( Mol Wt of Al = 26.98 g mol-1 )
Dividing by the stoichiometry of Al = 2.3165 / 2 = 1.158
no .of moles of Fe2O3 = 133 / 159.69 = 0.833 moles ( Mol Wt of Fe2O3 = 159.69 g mol-1 )
Dividing by the stoichiometry of Fe2O3 = 0.833 / 1 = 0.833
So, Fe2O3 will be the limiting reagent here because its value after the stated operation is less.
No . of moles of Al2O3 will be equal to no. of moles of Fe2O3 from the balanced equation = 0.833 mole
Wt of Al2O3 formed = 0.833 * 101.96 = 84.933 g ( Mol Wt of Al2O3 = 101.96 g mol-1 )
% Yield = (78.2 / 84.933) *100 = 92.07 %.
Thank You
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