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3 years ago

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The first step in the process used to recover zinc metal from zinc sulfide ore is the reaction of zinc sulfide with oxygen gas t

o produce zinc oxide and sulfur dioxide. 2ZnS(s) 3O2(g)⟶2ZnO(s) 2SO2(g) When the external pressure is 1.523×105 Pa and the temperature is 675 K, the amount of work performed is 850.1 J. Calculate how many grams of oxygen are consumed in the reaction.

1 answer:

Vikentia [17]3 years ago

4 0

Answer : The mass of oxygen consumed in the reaction is 14.6 grams.

Explanation :

First we have to calculate the change in moles of oxygen.

Formula used :

$w=\Delta nRT$

or,

$\Delta n=\frac{w}{RT}$

where,

w = work done = 850.1 J

$\Delta n$ = change in moles of gas = ?

R = gas constant = 8.314 J/mol.K

T = temperature = 675 K

Now put all the given values in the above formula, we get:

$\Delta n=\frac{850.1J}{(8.314J/mol.K)\times (675K)}=0.152mol$

Now we have to calculate the mass of oxygen consumed.

The balanced chemical reaction is,

$2ZnS(s)+3O_2(g)\rightarrow 2ZnO(s)+2SO_2(g)$

Here, the volume changes of solids can be ignored.

From the balanced chemical reaction, we conclude that

The number of moles oxygen required for the complete reaction = 3

Change in mole of oxygen = 0.152 mol

Molar mass of $O_2$ = 32 g/mol

So,

The mass of oxygen consumed = 3 x 0.152 mol x 32 g/mol = 14.6 g

Therefore, the mass of oxygen consumed in the reaction is 14.6 grams.

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Three complete orders on each side of the m=0 order can be produced in addition to the m = 0 order.

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3 years ago

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Many chemistry problems result in equations of the form

Gnom [1K]

Answer:

When this equation is solved, the two values of the unknown are 0.0643 and -0.082

Explanation:

Given

$1.77 * 10^{-2} = \frac{x^2}{0.298 - x}$ --- the actual equation

Required

The values of x

We have:

$1.77 * 10^{-2} = \frac{x^2}{0.298 - x}$

Cross Multiply

$1.77 * 10^{-2} * (0.298 - x)= x^2$

Multiply both sides by 100

$1.77 * (0.298 - x)= 100x^2$

Open bracket

$0.52746 - 1.77x= 100x^2$

Rewrite as:

$100x^2 + 1.77x - 0.52746 =0$

Using quadratic formula:

$x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where:

$a = 100; b = 1.77; c = -0.52746$

So, we have:

$x = \frac{-1.77 \± \sqrt{1.77^2 - 4*100*- 0.52746 }}{2*100}$

$x = \frac{-1.77 \± \sqrt{214.1169}}{2*100}$

$x = \frac{-1.77 \± 14.63}{200}$

Split

$x = \frac{-1.77 + 14.63}{200}\ or\ x = \frac{-1.77 - 14.63}{200}$

$x = \frac{12.86}{200}\ or\ x = \frac{-16.40}{200}$

$x = 0.0643\ or\ x = -0.082$

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3 years ago

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3 years ago

Hydrazine (N2H4) is used as rocket fuel. It reacts with oxygen to form nitrogen and water.

Marina86 [1]

Answer:

See explanation below for answers

Explanation:

This is a stochiometry reaction. LEt's write the overall reaction again:

N₂H₄ + O₂ ---------> N₂ + 2H₂O

This reaction is taking place at Standard temperature and pressure conditions (STP) which are P = 1 atm and T = 273 K. To know the volume of N₂ formed, we need to know first how many moles are formed, and this can be calculated with the reagents and the limiting reagent. Let's calculate the moles first of the reagents:

MM N₂H₄ = 32 g/mol; MM O₂ = 32 g/mol

mol N₂H₄ = 2000 / 32 = 62.5 moles

mol O₂ ? 2100 / 32 = 65.63 moles

Now that we have the moles, we need to apply the stochiometry and calculate the limiting reagent. According to the overall reaction we have a mole ratio of 1:1 between N₂H₄ and O₂, therefore:

1 mole N₂H₄ ---------> 1 mole O₂

62.5 moles ----------> X

X = 62.5 moles of O₂

But we have 65.63 moles, therefore, the limiting reactant is the N₂H₄.

We also have a 1:1 mole ratio with the N₂, so:

moles N₂H₄ = moles N₂ = 62.5 moles

Now that we have the moles, we can calculate the volume with the ideal gas equation:

PV = nRT

V = nRT / P

R: gas constant (0.082 L atm / K mol)

Replacing we have:

v = 62.5 * 0.082 * 273 / 1

V = 1399.13 L of N₂

Now, how many grams of the excess remains?, we know how many moles are reacting so, let's see how much is left:

moles remaining = 65.63 - 62.5 = 3.12 moles

then the mass of oxygen:

m = 3.12 * 32 = 100.16 g of O₂

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3 years ago

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