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Zolol [24]
3 years ago
9

How many moles of O2 are needed to react with 2.35 mol of C2H2?

Chemistry
2 answers:
adelina 88 [10]3 years ago
7 0

Answer : The number of moles of O_2 needed are 5.88 moles.

Solution : Given,

Moles of C_2H_2 = 2.35 mole

The balanced chemical reaction will be,

2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 moles of C_2H_2 react with 5 moles of O_2

So, 2.35 moles of C_2H_2 react with \frac{2.35}{2}\times 5=5.88 moles of O_2

Therefore, the number of moles of O_2 needed are 5.88 moles.

Mrrafil [7]3 years ago
5 0
2 C2H2 + 5 O2 --> 4 CO2 + 2 H2O

2.35 mol C2H2 - x mol O2
2 mol C2H2 - 5 mol O2

x =  \frac{2.35 \times 5}{2}  = 5.875 \: mol
answer: 5.875 mol
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If you are given an ideal gas with pressure (p)259,392.00 pa and temperature (T)=200°c of 1 mole Argon gas in a volume 8.8dm3,ca
GuDViN [60]

Answer: R=4.82436 \frac{Pa. m^{3}}{mol. K}

Explanation:

The Ideal Gas equation is:  

P.V=n.R.T  (1)

Where:  

P is the pressure of the gas  

n the number of moles of gas  

R=8.3144598 \frac{Pa. m^{3}}{mol. K} is the gas constant  

T is the absolute temperature of the gas in Kelvin.

V is the volume

It is important to note that the behavior of a real gas is far from that of an ideal gas, taking into account that <u>an ideal gas is a single hypothetical gas</u>. However, under specific conditions of standard temperature and pressure (T=0\°C=273.15 K and P=1 atm=101,3 kPa) one mole of real gas (especially in noble gases such as Argon) will behave like an ideal gas and the constant R will be 8.3144598 \frac{Pa. m^{3}}{mol. K}.

However, in this case we are not working with standard temperature and pressure, therefore, even if we are working with Argon, the value of R will be far from the constant of the ideal gases.

Having this clarified, let's isolate R from (1):

R=\frac{PV}{nT}  (2)

Where:

P=259392 Pa

n=1 mole

T=200\°C=473.15 K is the absolute temperature of the gas in Kelvin.

V=8.8 dm^{3}=0.0088 m^{3}

R=\frac{(259392 Pa)(0.0088 m^{3})}{(1 mole)(473.15 K)}  (3)

Finally:

R=4.82436 \frac{Pa. m^{3}}{mol. K}  

4 0
3 years ago
g Green plants use light from the Sun to drive photosynthesis. Photosynthesis is a chemical reaction in which water and carbon d
LenaWriter [7]

Answer:

6.49g

Explanation:

Let's consider the balanced reaction for photosynthesis.

6 CO₂ + 6 H₂O = C₆H₁₂O₆ + 6 O₂

We can establish the following relations:

  • 1 mole of CO₂ has a mass of 44.01 g (MW 44.01)
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The mass of glucose produced by the reaction of 9.51 g of carbon dioxide is:

9.51gCO_2 \times \frac{1molCO_2}{44.01gCO_2}  \times \frac{1molGlucose}{6molCO_2}  \times \frac{180.16gGlucose}{1molGlucose} = 6.49g Glucose

6 0
3 years ago
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The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product str
iren2701 [21]

Answer:

Explanation:

The first thing to do is to write out the chemical equation showing the reaction between the two chemical compounds.

C2H4 + HBr ------------------------> C2H5Br.

The chemical species/compounds reacted in the ratio of 1 : 1 : 1. That is one mole of C2H4 reacted with one mole of HBr to give one mole of C2H5Br.

So, we are given; C2H5Br = 58.7 mol%, HBr = 18.3 mol%. Thus, for C2H4, we have [ 100 - (58.7 + 18.3)] = 23 mol%. Therefore, the limiting reagent is HBr.

So, if we have 100 mol/s ;

C2H5Br = 100 × 0.587 = 58.7 mol/s, HBr = 100 × 0.183 = 18.3 mol/s and C2H4 = 100 × 0.23 = 23 mol/s.

Therefore, the total value of the flow rate in the feed = (HBr in feed) mol/s + (C2H4 in feed) mol/s = (58.7 + 18.3)mol/s + (58.7 + 23) mol/s =( 77 + 81.7) mol/s = 158.7 mol/s. .

So, 100/158.7 × 30.5 = 192.2.

Also the feed composition in HBr and C2H4 is;

HBr = 77/158.7 = 0.4852, C2H4 = 81.7/158.7 = 0.515.

Therefore, the number of HBr in feed and C2H4 in feed = 160m/s and 200m/s respectively (approx.)

Therefore, the fraction conversion is;

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2 years ago
Matt's cube, after 5 trial, had an average destiny of 7.40. g/cm what is it made of ?​
alekssr [168]

Explanation:

5 trial, had an average destiny of 7.40. g/cm

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Leona [35]

Answer:

There is None

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This is because it is a derived function dependent on other factors.

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3 years ago
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