The illustration to the left represents a mixture of oxygen (red) and fluorine (
ID: 1051814 • Letter: T
Question
The illustration to the left represents a mixture of oxygen (red) and fluorine (green) molecules. If the molecules in the above illustration react to form OF_2 according to the equation O_2 + 2 F_2 2 OF_2, For the following reaction, 23.3 grams of sodium chloride are allowed to react with 54.5 gram* ofsilyee nitrate sodium chloride(aq) + silver nitrate(aq) silver chloride(s) + sodium nitrate(aq) What is the maximum amount of silver chloride that can be formed? (Answer here) grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams For the following reaction. 19.0 grams of iron are allowed to react with 9.18 grams oxygen gas, iron(s) + oxygen(g) iron(II) oxide(s) What is the maximum mass of iron(II) oxide that can be formed? (ANSWHR HERE) grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction complete? (ANSWER HERE) grams.Explanation / Answer
Problem two:
NaCl(aq) + AgNO3(aq) -----------------------------> NaNO3(aq) + AgCl(s)
From the balanced chemical equation we get the information that:
58.5 g of NaCl is reacting with--------170g of AgNO3 -----------forms 143.5g AgCl
Then 23.3 g of NaCl reacts with ------------? g of AgNO3
= 23.3 * 170/ 58.5
= 67.7g of AgNO3.
Hence 23.3g of NaCl reacts with 67.7g of AgNO3.
But in the question it is given that we are taking only 54.5g of AgNO3.
Hence the amount of AgNO3 required to react with 23.3g of NaCl is less than required.
Hence AgNO3 is the limiting reagent.
170g of AgNO3 -----------forms 143.5g AgCl :this is according to the balanced equation.
Then 54.5g of AgNO3 --------- forms ? grams of AgCl
= 54.5g x 143.5g / 170g
= 46g of AgCl is the maximum amount formed.
The formula of limiting reagent is AgNO3.
The amount of excess reagent remained can be calculated as below:
NaCl is the excess reagent.
From the balanced chemial equation we get the information that :
58.5 g of NaCl is reacting with--------170g of AgNO3
then ? g of NaCl is required to react with------ 54.5g of AgNO3
= 54.5g X 58.5 g / 170g
= 18.75g of NaCl is required to react with 54.5g of AgNO3.
Excess amount remained = 23.3g - 18.75g
= 4.55g of NaCl
Problem three:
4 Fe(s) + 3O2 (g) ----------------------->2 Fe2O3(s)
From the balanced chemical equation we get the information that:
224g of Fe is reacting with--------- 96g of O2 and forms----------- 320g of Fe2O3.
Then 19g of Fe reacts with -----------? g of O2
= 19g x 96g / 224g
= 8.14g of O2 is required.
In the question it is given that we are taking 9.18g of O2.
So it is more than required, hence O2 is the excess reagent.
Fe is the limiting reagent.
The amount of excess reagent remained after the reaction is : 9.18g - 8.14g = 1.04g of O2 remains as excess.
The formula of the limiting reagent is Fe(s)
The maximum mass of Fe2O3 formed can be calculated as below:
From the balanced chemical equation we get the information that:
224g of Fe forms------------------ 320g of Fe2O3
then 19g of Fe forms---------------------? g of Fe2O3
= 27.14g of Fe2O3 is formed.
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