he reported molar absorptivity of Red Dye #3 is 1.217 L mol-1 cm-1. If a solution of Red Dye #3 displays an absorbance of 0.699
1 answer:
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The image is not given in the question, it is attached below:
<u>Answer:</u> The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.
<u>Explanation:</u>
In the given image:
Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms
The combination of 1 black and 2 red spheres will represent carbon dioxide compound
The combination of 2 blue spheres will represent nitrogen molecule
The combination of 1 blue and 1 red sphere will represent nitrogen monoxide compound
The combination of 1 black and 1 red sphere will represent nitrogen monoxide compound
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
We are given:
Given moles of NO = 6 moles
Given moles of CO = 4 moles
For the given chemical equation:
By stoichiometry of the reaction:
If 2 moles of CO reacts with 2 moles of NO
So, 4 moles of CO will react with = of NO
As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, CO is considered a limiting reagent because it limits the formation of the product.
Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose =
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution =
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution =
Volume of the solution taken =
Molarity of the solution after dilution =
Volume of the solution after dilution=
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose =
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.