Can somebody help me plz

How many moles of potassium hydroxide are needed to completely react with 2.94 moles of aluminum sulfate

ArbitrLikvidat [17]

Answer:- Third choice is correct, 17.6 moles

Solution:- The given balanced equation is:

Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4

We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.

From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.

It is a simple mole to mole conversion problem. We solve it using dimensional set up as:

2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})

= 17.6 mol KOH

So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.

6 0

3 years ago

In the reaction BaCO3 + 2HNO3 + Ba(NO3)2 + CO2 + H2O, what mass of Ba(NO3)2 can be formed by combining 55 g BaCO3 and 26 g HNO3

Nataliya [291]

From the stoichiometry of the reaction, the mass of barium nitrate produced is 54.9 g.

<h3>Stoichiometry</h3>

The term stoichiometry refers to mass - volume relationships. Stoichiometry can be used to calculate the amount, mass or volume of reactants and products from the balanced reaction equation.

The equation of the reaction is written as follows;

BaCO3 + 2HNO3 ------> Ba(NO3)2 + CO2 + H2O

Number of moles of BaCO3 = 55 g/197.34 g/mol = 0.28 moles

Number of moles of HNO3 = 26 g/63.01 g/mol = 0.41 moles

From the reaction equation;

1 mole of BaCO3 reacts with 2 moles of HNO3

0.28 moles of BaCO3 reacts with 0.28 moles × 2 moles/1 mole = 0.56 moles

There is not enough HNO3 hence it is the limiting reactant.

Number of moles of Ba(NO3)2 produced is obtained from;

2 moles of HNO3 yields 1 mole of Ba(NO3)2

0.41 moles of HNO3 yields 0.41 moles × 1 mole/2 moles

= 0.21 moles of Ba(NO3)2

Mass of Ba(NO3)2 = 0.21 moles × 261.337 g/mol = 54.9 g

Learn more about stoichiometry: brainly.com/question/9743981

8 0

2 years ago

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

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

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

2 years ago

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa

NaOH + CH3COOH → CH3COONa + H2O

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M

CH3COONa = 0.0076 / 0.071 = 0.1070M

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.

pH using Henderson - Hasselbalch equation:

pH = pKa + log ([salt]/[acid])

pH = 4.74 + log ( 0.1070/0.1493)

pH = 4.74 + log 0.717

pH = 4.74 + (-0.14)

pH = 4.60.

7 0

2 years ago

Magnesium reacts with a certain element to form a compound with the general formula MgX. What would the most likely formula be f

ziro4ka [17]

Answer:K2X

Explanation: Valency can be defined as the combining power of an element. It is the valency that dictates the value an element will have when writing a chemical formula for its compound.

MgX is a compound of magnesium and an element X. The valency of magnesium in most of its compound is +2. Now for the 2 to have been absent in the chemical formula, this shows that the element X itself have a valency if -2 for the valencies of both to have canceled out.

Now considering the element potassium, it is an alkaline metal belonging to group 1 of the periodic table. Hence, it is expected that it has a valency of +1

Forming a compound with element X means there would be an exchange of valencies between the two. We have established that x has a valency of -2. The formula of the compound thus formed by exchanging the valencies of both element would be K2X

7 0

2 years ago

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