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katen-ka-za [31]
3 years ago
5

Consider the five balanced chemical reactions listed below, all using O2 as a reactant. Normally, O2 is an excess reagent for re

actions because there is a large amount of it available in the air. For this question, suppose that you have5.73 moles of O2 and1.70 moles of any of the other reactants in every equation. Select those equations below in which O2 would be the limiting reactant.
A.C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)B.C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)C.H2S(g) + 2 O2(g) → SO3(g) + H2O(g)D.4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)E.None because in all the reactions O2 is in excess
Chemistry
1 answer:
ad-work [718]3 years ago
7 0

Answer: option E. None because in all the reactions O2 is in excess

Explanation:

You might be interested in
What is the predicted change in the boiling point of water when 1.50 g of
dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

⠀

\textsf {While} \:  \sf  {\Delta T_b}  \: \textsf{expression is used} \\  \textsf {for elevation of boiling point}

⠀

⠀

<u>The</u><u> </u><u>elevation</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>phenomenon</u><u> </u><u>in</u><u> </u><u>which</u><u> </u><u>there</u><u> </u><u>is</u><u> </u><u>increase</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>in</u><u> </u><u>solution</u><u>,</u><u> </u><u>when</u><u> </u><u>the</u><u> </u><u>particular</u><u> </u><u>type</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>is</u><u> </u><u>added</u><u> </u><u>to</u><u> </u><u>pure</u><u> </u><u>solvent</u><u>.</u>

⠀

⠀

\sf  \large \underline{The \:  formula \: to \:  be  \: used \:  in \:  this \:  question \:  is}  \\   \boxed{T_b = i \times  K_b \times  m}

⠀

⠀

Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

⠀

'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

⠀

'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

⠀

⠀

<u>To</u><u> </u><u>find</u><u> </u><u>molality</u><u>,</u><u> </u><u>we</u><u> </u><u>have</u><u> </u><u>to</u><u> </u><u>divide</u><u> </u><u>no</u><u>.</u><u> </u><u>of</u><u> </u><u>moles</u><u> </u><u>of</u><u> </u><u>solute</u><u> </u><u>by</u><u> </u><u>weight</u><u> </u><u>of</u><u> </u><u>solution</u>

⠀

While first we need to no. of moles

\sf \implies no. \: of \: moles =  \frac{weight \: of \: solute}{molar \: mass \: of \: solute}  \\  \\ \implies \sf no. \: of \: moles =  \frac{1.5}{208.23}  \\  \\  \sf \implies  no. \: of \: moles = 0.0072

⠀

⠀

<u>Now</u><u>,</u><u> </u><u>we</u><u> </u><u>will</u><u> </u><u>find</u><u> </u><u>molality</u>

⠀

\sf  \hookrightarrow molality =  \frac{no.\: of \: moles}{weight \: of \: solution}  \\  \\  \sf  \hookrightarrow molality =  \frac{0.072}{1.5}  \\  \\  \sf  \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}

⠀

⠀

\textsf{ \large{ \underline{Now substituting the required values}}}

⠀

\sf \longmapsto \Delta T_b = 3  \times 0.51  \times 0.0048 \\  \\ \\     \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}

⠀

⠀

⠀

<u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>change</u><u> </u><u>in</u><u> </u><u>boiling</u><u> </u><u>point</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>7</u><u>3</u><u>5</u><u>°</u><u>C</u><u>.</u>

7 0
1 year ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
Read 2 more answers
An element forms an oxide, E₂O₃, and a fluoride, EF₃.(b) How does the group to which E belongs affect the properties of the oxid
xz_007 [3.2K]

Of all the elements, fluorine is the most electronegative and reactive. Fluorine is a diatomic, pale yellow, extremely corrosive, combustible gas with a strong smell. The lightest halogen is it. It produces oxygen and the incredibly corrosive hydrofluoric acid when it combines strongly with water.

<h3>The properties of the oxide and the fluoride?</h3>
  • 1. A mixture of oxygen fluorides with an atomic ratio OF in the range of 1.1-2.04 is generated when fluorine and oxygen mixes are easily circulated through a silent electric discharge.
  • Depending on where you reside in the UK, fluoride is a naturally occurring mineral that is present in water in variable concentrations. It is added to many types of toothpaste and, in some locations, the water supply through a procedure known as fluoridation because it can aid in the prevention of tooth decay.
  • Fluoride stops tooth decay by strengthening the enamel's resistance to acid attack. They also quicken the process of good minerals accumulating in the enamel, further delaying the onset of deterioration. Studies also suggest that fluoride may occasionally be able to stop tooth decay that has already begun.

To know more about Fluoride please click here : brainly.com/question/10929330

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6 0
1 year ago
What is waste good for?
harkovskaia [24]
A because that honestly just makes the most sense
6 0
2 years ago
Find ΔG° for the reactions in Problem 20.51 using ΔHf° and S° values.
pashok25 [27]

The value of ΔG° (gibbs free energy change)of the given reaction is -1131.94kj .

Given ,

Balanced chemical equation is given by,

2Mg(s) + O2 (g) → 2MgO (s)

Here , the standard enthalpy of Mg  and O2 gas are zero because they are in the most stable state .

thus , the value of ΔG°rxn is given by ,

ΔG°rxn = ΔG°(products ) - ΔG°(reactants )

ΔG°rxn = [2×(-565.97 )] - 0

ΔG°rxn = 2×(-565.97)

ΔG°rxn = -1131.94kj

Hence , the value of ΔG° of the given reaction is -1131.94kj .

<h3>What is a balance chemical reaction ?</h3>

A balanced chemical reaction is a type of reaction which include the reactants and products in same amount i.e. the no of mole on both side of the reaction is remains same or balanced .

Learn more about chemical reaction here :

brainly.com/question/1893305

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Disclaimer :

incomplete question . here is the complete question .

Question :

Calculate ΔG° for the reaction using ΔG°f values :

2Mg(s) + O2 (g) → 2MgO (s)

Find the ΔG° for the reaction using ΔH°f and S° values .

3 0
1 year ago
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