Chemistry HW..limiting reactant?
If a peice of Na metal (mass 8.20g) is burned in a jar that contains 20.62g Cl2 gas what is the limiting reactant?
How many moles of NaCl have been formed?
How many grams of excess reactant remain unreacted in the jar after the reaction is completed?
The balanced chemical equation:
2 Na (s) + Cl2 (g) --> 2 NaCl (s)
Thank you so much for your help!!
Answer:
First convert your reactants to moles by dividing them by their respective molecular weights.
This gives us
.35 moles of Na and
.29 moles of Cl2
looking at the Stochiometry, if all .29 moles of Cl2 were used up, we would get .58 moles of product (because 1Cl2 contributes to 2NaCls).
If all .35 moles of Na were used up, we would only get .35 moles of product (because 1Na contributes to 1NaCl). This is the limiting reagent and the Cl2 is in excess.
Because we have a one to one ratio of Na and NaCl, we will make as many moles of NaCl as we have Na. So .35 moles.
If .35 moles of Na were used up, then we used half as many moles of Cl2, so we used .175 moles of Cl2, which means that .115 moles of Cl2 are left unreacted, then convert that back to grams, 8.15grams (not exact of course, because I did not use exact values)
According to the equation, one gram atom of sodium reacts with one gram atom of chlorine to form one mole of sodium chloride.
You have 8.2/23 gram atoms of sodium = 0.357
You have 20.62/35.5 gram atoms of chlorine = .581
Sodium is the limiting reagent.
You will form 0.357 moles of sodium chloride or
0.357 * 58.5 =20.9 grams
.357 gram atoms of chlorine will react leaving 0.581- 0.357 gram atoms of chlorine = 0.224 gram atoms of chlorine
.224 gram atoms of chlorine = .224*35.5= 7.95 grams of unreacted chlorine left over
The limiting reagent from the equation is sodium (Na)
To determine this you divide the masses by their relative atomic masses.
From the above the relative atomic masses of Na and Cl2 are 23.0 and 35.5respectively. When you do the calculation, the smaller value is the limiting reagent
The vallues would be 0.3(Na) and 0.5(Cl2)
1. first get the number of moles of Na and Cl using their atomic weights
2. multiply that by the mass of your Na and Cl
8.20g (1 mole Na/22.99) = 0.3567 moles of Na; for Cl it's 20.62 g (1mole Cl/35.45) = 0.582 moles of Cl
3. compare0.3567 Na and 0.582 Cl, limiting reactant means if that element is used up the reaction ceases (of course you know that). Now this is where you take in consideration the balanced equation. Take note that you've got 2 molecules there of Na.
I think you can continue from here since I already gave you enough info to finish your homework.
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How many moles of NaCl have been formed?
How many grams of excess reactant remain unreacted in the jar after the reaction is completed?
The balanced chemical equation:
2 Na (s) + Cl2 (g) --> 2 NaCl (s)
Thank you so much for your help!!
Answer:
First convert your reactants to moles by dividing them by their respective molecular weights.
This gives us
.35 moles of Na and
.29 moles of Cl2
looking at the Stochiometry, if all .29 moles of Cl2 were used up, we would get .58 moles of product (because 1Cl2 contributes to 2NaCls).
If all .35 moles of Na were used up, we would only get .35 moles of product (because 1Na contributes to 1NaCl). This is the limiting reagent and the Cl2 is in excess.
Because we have a one to one ratio of Na and NaCl, we will make as many moles of NaCl as we have Na. So .35 moles.
If .35 moles of Na were used up, then we used half as many moles of Cl2, so we used .175 moles of Cl2, which means that .115 moles of Cl2 are left unreacted, then convert that back to grams, 8.15grams (not exact of course, because I did not use exact values)
According to the equation, one gram atom of sodium reacts with one gram atom of chlorine to form one mole of sodium chloride.
You have 8.2/23 gram atoms of sodium = 0.357
You have 20.62/35.5 gram atoms of chlorine = .581
Sodium is the limiting reagent.
You will form 0.357 moles of sodium chloride or
0.357 * 58.5 =20.9 grams
.357 gram atoms of chlorine will react leaving 0.581- 0.357 gram atoms of chlorine = 0.224 gram atoms of chlorine
.224 gram atoms of chlorine = .224*35.5= 7.95 grams of unreacted chlorine left over
The limiting reagent from the equation is sodium (Na)
To determine this you divide the masses by their relative atomic masses.
From the above the relative atomic masses of Na and Cl2 are 23.0 and 35.5respectively. When you do the calculation, the smaller value is the limiting reagent
The vallues would be 0.3(Na) and 0.5(Cl2)
1. first get the number of moles of Na and Cl using their atomic weights
2. multiply that by the mass of your Na and Cl
8.20g (1 mole Na/22.99) = 0.3567 moles of Na; for Cl it's 20.62 g (1mole Cl/35.45) = 0.582 moles of Cl
3. compare0.3567 Na and 0.582 Cl, limiting reactant means if that element is used up the reaction ceases (of course you know that). Now this is where you take in consideration the balanced equation. Take note that you've got 2 molecules there of Na.
I think you can continue from here since I already gave you enough info to finish your homework.
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