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3 years ago

\ce{Zn + CuCl2 -> ZnCl2 + Cu}Zn+CuCl2 ZnCl2 +Cu

Chemistry

1 answer:

Kruka [31]3 years ago

3 0

Answer:

0.321 mole of ZnCl₂ have been produced from 21 g of Zn

Explanation:

The reaction is this one:

Zn (s) + CuCl₂ (aq) → ZnCl₂ (aq) + Cu(s)

A redox one.

If the CuCl₂ is available in excess, we consider the limiting reactant, Zn.

Molar mass Zn = 65.38 g/m

Mass / Molar mass = Mol

21 g/ 65.38 g/m = 0.321 mole

As ratio is 1:1, I will produce the same amount of mole, so I'll make 0.321 mole of ZnCl₂

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Please help me! I don't understand this at all. All the info is in the picture. Thank you so much!!!

kykrilka [37]

Answer:

H₂S + Cl₂ —> S + 2HCl

Explanation:

? + Cl₂ —> S + 2HCl

To balance the equation above, we must recognise what atoms are present in the products.

The products contains S, H and Cl.

Thus, S, H and Cl must also be present in the reactants.

Considering the equation given above, we can see clearly that H and S is missing in the reactants.

H and S together as a compound is expressed as H₂S.

Now, we shall input H₂S into the equation to obtain the complete equation. This is illustrated below:

? + Cl₂ —> S + 2HCl

H₂S + Cl₂ —> S + 2HCl

Next, we shall verify to see if the equation is balanced.

There are 2 atoms of H on both sides of the equation.

There are 2 atoms of Cl on both sides of the equation.

1 atom of S exist on both sides of the equation.

Thus, the equation is balanced.

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3 years ago

Which of the following is a product of photosynthesis? Carbon dioxide Glucose Water Xylem

serious [3.7K]

The answer is glucose.
Hope it helps!

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3 years ago

Question 7

Paladinen [302]

Answer:

2HCI + Na2S - > H2S + 2NaCl

Explanation:

2HCI + Na2S -> H2S + 2NaCl

before reaction: 2H, 2Cl, 2Na, 1S

after reaction: 2H, 2Cl, 2Na, 1S

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4 years ago

Organisms that are made up of prokaryotic cells are NOT ____________.

Ulleksa [173]

Answer:

Multicellular

Explanation:

Prokaryotic cells are unicellular because the do not have a nucleus and lack organelles, while eukaryotic cells do have a nucleus, have organelles, and are are multicellular. :)

6 0

3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

\ce{Zn + CuCl2 -> ZnCl2 + Cu}Zn+CuCl2 ​ ​ ZnCl2 ​ +Cu

\ce{Zn + CuCl2 -> ZnCl2 + Cu}Zn+CuCl2 ZnCl2 +Cu