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3 years ago

Calculate the moles of SO₂ produced when 5 moles of FeS₂ reacts according to the equation: 4FeS₂+ 11O₂ → 2Fe₂O₃ + 8SO₂

Chemistry

1 answer:

Gala2k [10]3 years ago

7 0

Answer:

10 moles of SO₂ are produced when 5 moles of FeS₂

Explanation:

Stoichiometry: it is the theoretical proportion in which the chemical species are combined in a chemical reaction. The stoichiometric equation of a chemical reaction relates molecules or number of moles of all the reagents and products that participate in the reaction.

In other words, stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships established are molar relationships (that is, moles) between the compounds or elements that make up the chemical equation.

The stoichiometric coefficients of a chemical reaction indicate the proportion in which said substances react.

Taking into account the above, you can apply the following rule of three: by stoichiometry if 4 moles of FeS₂ produce 8 moles of SO₂, then when reacting 5 moles of FeS₂ how many moles of SO₂ will they produce?

$moles of SO_{2} =\frac{5 moles of FeS_{2} *8moles of SO_{2}}{4 moles of FeS_{2}}$

moles of SO₂= 10

10 moles of SO₂ are produced when 5 moles of FeS₂

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1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

Please help!

STALIN [3.7K]

Pitch is sometimes defined as the fundamental frequency of a sound wave. For most practical purposes, this is fine, and pitch and frequency can be thought of as equivalent. On the other hand, for most practical purposes, amplitude can be thought of as volume.

However, technically, pitch and volume are human perceptions. Thus, our perception of pitch and volume are not solely based on frequency and amplitude respectively, but are based on a combination of both. Frequency overwhelming dictates perceived pitch, but amplitude also does have some small, small effect on our pitch perception, especially when it is very large. For example, a very loud sound can have a different perceived pitch than you would predict from its frequency alone.

Hope that helps!

6 0

3 years ago

Necesito ayuda porfavor

AlexFokin [52]

Answer:

thank for the points ❤️

Explanation:

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8 0

3 years ago

What would be the change in pressure in a sealed 10.0 l vessel due to the formation of n2 gas when the ammonium nitrite in 2.40

larisa86 [58]

The change in pressure in a sealed 10.0L vessel is 5.28 atm

calculation

The pressure is calculated using the ideal gas equation

That is P=n RT

where;

P (pressure)= ?

v( volume) = 10.0 L

n( number of moles) which is calculated as below

write the equation for decomposition of NH₄NO₂

NH₄NO₂ → N₂ +2H₂O

Find the moles of NH₄NO₂

moles = molarity x volume in liters

= 2.40 l x 0.900 M =2.16 moles

Use the mole ratio to determine the moles of N₂

that is from equation above NH₄NO₂:N₂ is 1:1 therefore the moles of N₂ is also =2.16 moles

R(gas constant) =0.0821 l.atm/mol.K

T(temperature) = 25° c into kelvin = 25 +273 =298 K

make p the subject of the formula by diving both side by V

P = nRT/V

p ={ (2.16 moles x 0.0821 L.atm/mol.K x 298 K) /10.0 L} = 5.28 atm.

3 0

2 years ago

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What two things must occur in order for a light ray to refract when it strikes

Ann [662]

Answer:

When light passes from one transparent medium to another, the rays are bent toward the surface normal if the speed of light is smaller in the second medium than in the first. The rays are bent away from this normal if the speed of light in the second medium is greater than in the first.

Explanation:

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2 years ago