Consider the reaction
2Al + 3CuSO4 = Al2(SO4)3 + 3Cu
What is the maximum amount of Cu (63.5
g/mol) that could be produced by reacting
20.0 grams of Al (27.0 g/mol) with excess
CuSO4?
1. 70.6 g
2. 57.2 g
3. 129 g
4. 31.4 g
Consider the reaction
4As + 3O2 = 2 As 2O3
What is the maximum amount of As2O3
which could be produced by reacting 10.0
grams of As with excess oxygen?
1. 13.2 g
2. 6.6 g
3. 26.4 g
4. 2 g
5. 5 g
For the reaction
4NH3 + 5O2 = 4NO + 6H2O
how much water (18.0 g/mol) can be pro-duced from the reaction of 37.1 g of ammonia (NH3 at 17.0 g/mol), with an excess of oxy-
gen?
Answer in units of g.
Laughing gas (nitrous oxide, N2O) is some-
times used as an anesthetic in dentistry. It is
produced when ammonium nitrate is decom-
posed according to the reaction
NH4NO3(s) = N2O(g) + 2H2O(§¤) .
How much water is produced if 43.5 g of
N2O is produced in this reaction?
Answer in units of g.
For the reaction
?P4 +? I2 !?PI3 ,
what is the maximum amount of PI3
(411.687 g/mol) which could be formed from
10.39 g of P4 (123.895 g/mol) and 16.17 g of
I2 (253.809 g/mol)?
Answer in units of g.
70.6
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WHAT IS THE MAXIMUM AMOUNT OF cR2S3 IN GRAMS OBTAINABLE
To find the maximum amount of product produced in a chemical reaction, you need to use stoichiometry and calculate the amount of each reactant in moles, and then determine which reactant is limiting. The reactant that is completely consumed and limits the amount of product that can be formed is known as the limiting reactant.
Let's go through each of the given reactions and calculate the maximum amount of product formed:
1) Reaction: 2Al + 3CuSO4 = Al2(SO4)3 + 3Cu
To find the limiting reactant, you need to compare the number of moles of aluminum (Al) and copper sulfate (CuSO4).
- First, calculate the number of moles of Al:
20.0 g Al * (1 mol Al / 27.0 g Al) = 0.741 mol Al
- Next, calculate the number of moles of CuSO4:
Since CuSO4 is in excess, we assume all of it will react, so we don't need to calculate the moles.
According to the stoichiometry of the balanced equation, the ratio of Al to Cu is 2:3. This means that 2 moles of Al react with 3 moles of CuSO4 to produce 3 moles of Cu. Therefore, 0.741 mol Al would react with (0.741 mol Al * 3 mol Cu / 2 mol Al) = 1.1125 mol Cu.
- Finally, calculate the mass of Cu:
1.1125 mol Cu * (63.5 g Cu / 1 mol Cu) = 70.684 g Cu
The maximum amount of Cu that could be produced is approximately 70.684 g (option 1).
2) Reaction: 4As + 3O2 = 2As2O3
To find the limiting reactant, you need to compare the number of moles of As and O2.
- First, calculate the number of moles of As:
10.0 g As * (1 mol As / 74.92 g As) = 0.1336 mol As
- Next, calculate the number of moles of O2:
Since O2 is in excess, we assume all of it will react, so we don't need to calculate the moles.
According to the stoichiometry of the balanced equation, the ratio of As to As2O3 is 4:2. This means that 4 moles of As react with 3 moles of O2 to produce 2 moles of As2O3. Therefore, 0.1336 mol As would react with (0.1336 mol As * 2 mol As2O3 / 4 mol As) = 0.0668 mol As2O3.
- Finally, calculate the mass of As2O3:
0.0668 mol As2O3 * (197.84 g As2O3 / 1 mol As2O3) = 13.229 g As2O3
The maximum amount of As2O3 that could be produced is approximately 13.229 g (option 1).
3) Reaction: 4NH3 + 5O2 = 4NO + 6H2O
Since the question asks for the amount of water produced, we need to find the limiting reactant by comparing the moles of NH3 and O2.
- First, calculate the number of moles of NH3:
37.1 g NH3 * (1 mol NH3 / 17.0 g NH3) = 2.18 mol NH3
- Next, calculate the number of moles of O2:
Since O2 is in excess, we assume all of it will react, so we don't need to calculate the moles.
According to the stoichiometry of the balanced equation, the ratio of NH3 to H2O is 4:6. This means that 4 moles of NH3 react with 5 moles of O2 to produce 6 moles of H2O. Therefore, 2.18 mol NH3 would react with (2.18 mol NH3 * 6 mol H2O / 4 mol NH3) = 3.27 mol H2O.
- Finally, calculate the mass of H2O:
3.27 mol H2O * (18.0 g H2O / 1 mol H2O) = 58.86 g H2O
The maximum amount of water that could be produced is approximately 58.86 g.
4) Reaction: NH4NO3(s) = N2O(g) + 2H2O(l)
To find the amount of water produced, we need to use the given amount of N2O.
- Calculate the number of moles of N2O:
43.5 g N2O * (1 mol N2O / 44.02 g N2O) = 0.988 mol N2O
According to the stoichiometry of the balanced equation, the ratio of N2O to H2O is 1:2. This means that 1 mole of N2O reacts to produce 2 moles of H2O. Therefore, 0.988 mol N2O would produce (0.988 mol N2O * 2 mol H2O / 1 mol N2O) = 1.976 mol H2O.
- Finally, calculate the mass of H2O:
1.976 mol H2O * (18.0 g H2O / 1 mol H2O) = 35.568 g H2O
The amount of water produced is approximately 35.568 g.
5) Reaction: P4 + 5I2 = 4PI3
To find the limiting reactant, you need to compare the number of moles of P4 and I2.
- First, calculate the number of moles of P4:
10.39 g P4 * (1 mol P4 / 123.895 g P4) = 0.0839 mol P4
- Next, calculate the number of moles of I2:
16.17 g I2 * (1 mol I2 / 253.809 g I2) = 0.0638 mol I2
According to the stoichiometry of the balanced equation, the ratio of P4 to PI3 is 1:4. This means that 1 mole of P4 reacts with 5 moles of I2, producing 4 moles of PI3. Therefore, 0.0839 mol P4 would react with (0.0839 mol P4 * 4 mol PI3 / 1 mol P4) = 0.3356 mol PI3.
However, we can see that the limiting reactant is I2 because it has the smaller number of moles. Therefore, we need to calculate the amount of PI3 produced using the moles of I2.
- Finally, calculate the mass of PI3:
0.0638 mol I2 * (4 mol PI3 / 5 mol I2) * (411.687 g PI3 / 1 mol PI3) = 13.191 g PI3
The maximum amount of PI3 that could be formed is approximately 13.191 g (option 1).