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katrin2010 [14]
2 years ago
5

When 12 moles of o2 react with 1.1 mole of c10h8 what is the limiting reactant?

Chemistry
2 answers:
Trava [24]2 years ago
6 0
1.1   moles   of C10H8  is the  limiting  reagent  in   the  reaction between   reaction  C10H8   and   O2
.
C10H8  +  12O2  ---->  10CO2   + 4H2O

C10H8  is  the  limiting   reagent   since   1.1  moles  of  C10H8  is  totally   consumed   during  the  reaction
mariarad [96]2 years ago
6 0

Answer: C10H8 is the limiting Reactant

Explanation:

Ok, before delving into this question let us take some definitions.

WHAT IS A LIMITING REACTANT?

A limiting Reactant is also called a limiting reagent. From the English word ''limiting' that is setting a limit, a boundary to something. Thus, in Chemistry, a limiting reagent can be defined as the chemical substance (in a chemical reaction) that determines how much a chemical reaction can go on to generate product(s).

What is an Excess Reactant: an Excess Reactant is the excessive reactants in any chemical equation. Let just say, it is the opposite of a limiting reagent.

With all these been said, let us write down the balanced chemical equation of the reaction;

C10H8 + 12 O2 --------> 10 CO2 + 4H2O.

From the Chemical reaction above, we can see that one mole of C10H8 reacts with 10 moles of Oxygen,O2 to give 10 moles of Carbondioxide, CO2 and 4 moles of water,H2O.

After using 1 mole From C10H8 the reaction should likely come to an end. If one mole of C10H8 reacts completely with one mole of O2, means that there will be an EXCESS OF 11 moles. Thus, the LIMITING REAGENT/REACTANT IS C10H8.

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When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro
Bas_tet [7]
<span>Answer: For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees. 4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ. Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work. To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3. .0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
7 0
2 years ago
If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T
zhuklara [117]

Answer: The actual yield of B_2O_3 is 60.0 g

Explanation:-

The balanced chemical reaction :

B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)

Mass of B_2H_6 =Density\times Volume=1.131g/ml\times 36.9ml=41.7g

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles

According to stoichiometry:

1 mole of B_2H_6 gives = 1 mole of B_2O_3

1.51 moles of B_2H_6 gives =\frac{1}{1}\times 1.51=1.51 moles of B_2O_3

Theoretical yield of B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g

Percent yield of B_2O_3= 75.7\%

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

75.7\%=\frac{\text{Actual yield}}{79.3}\times 100

{\text{Actual yield}}=60.0g

Thus the actual yield of B_2O_3 is 60.0 g

7 0
2 years ago
Calculate the empirical formula for each of the following substances. (Express answer as a chemical formula) 1) 2.90 g of Ag and
exis [7]

Answer:

1) Ag3N

2)Na2S

3)NaHSO4

4) KNO3

Explanation:

We divide each mass by the element's relative atomic mass

1) 2.90/108-Ag, 0.125/14-N

0.027-Ag, 0.0089-N

Divide by the lowest ratio

0.027/0.0089-Ag, 0.0089/0.0089 N

3-Ag, 1-N

Empirical formula- Ag3N

2)2.22/23-Na, 1.55/32-S

0.097-Na, 0.048-S

Divide by the lowest ratio

0.097/0.048-Na, 0.048/0.048-S

2-Na, 1-S

Empirical formula- Na2S

3) 2.11/23-Na, 0.0900/1-H, 2.94/32-S,5.86/16-O

0.09-Na, 0.09-H, 0.09-S,0.366-O

Divide by the lowest ratio

0.09/0.09-Na, 0.09/0.09-H, 0.09/0.09-S, 0.366/0.09-O

1-Na, 1-H, 1-S, 4-O

Empirical formula- NaHSO4

4)1.84/39, 0.657/14-N, 2.25/16-O

0.047-K, 0.047-N, 0.14-O

Divide through by the lowest ratio

0.047/0.047-K, 0.047/0.047-N, 0.14/0.047-O

1-K, 1-N, O-3

Empirical formula- KNO3

4 0
2 years ago
What mass of solid lead would displace exactly 234.6 liters of water?
eimsori [14]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

The mass of solid lead would displace exactly 234.6 liters of water should be <span>2,674,440</span>
6 0
2 years ago
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The equilibrium constant for the reaction sr(s) + mg2+(aq) ⇌ sr2+(aq) + mg(s) is 2.69 × 1012 at 25°c. calculate e o for a cell m
Mariulka [41]
Sr(s)+Mg²+(aq)→Sr²+(aq)+Mg(s)
Number of e-'s transfered are, n=2. Equilibrium constant,
K=2.69×10∧12
ΔG=-2.303RT logK
R=gasconstant=8.314J/mol-k
T= temperature in K= 25 oC=25+273=298K
The value we get ΔG = -70922.3J. But ΔG = -nFE
n= number of e-'s transfered in the reaction =2
F= farady = 96500C
E=potential of the cell is what?
∴E = ΔG.nF
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8 0
2 years ago
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