How to balance this equation

11.9 g Cl2 is reacted with 10.7 g NaOH. How many moles of NaCl are produced?

melisa1 [442]

Answer:

1. 7.256g of NaCl

2. 47.33g of Cl2

Explanation:

2 moles of Na reacts to produce 2 moles of NaCl

8 moles of Na will still produce 8 moles of NaCl

Mass of NaCl = molar mass of Nacl/moles of Nacl

=58.5/8

=7.256g of NaCl

From the equation, 2 moles of Na reacts with 1 mole of Cl2

3/2 moles of Cl2 will react with 3 moles of Na

Mass of Cl2 = 71/1.5

=47.33g of Cl2

Explanation:

4 0

2 years ago

Determine the number of atoms that are in 1.15 mol of Cl 2.

Brums [2.3K]

Answer:

Explanation: I think its 4.91 x 10^25. Im not very sure, i just multipled 1.15 mol by the molar mass of Cl 2, which was 70.9 g. Then I multiplied that by avogadro's number. sorry if im wrong

4 0

2 years ago

Directions: Each set of lettered choices below refers to the numbered statements immediately following it. Select the one letter

maksim [4K]

Answer : The only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

Explanation :

Disproportionation reaction : It is defined as the reaction in which the same reactant undergoes both oxidation and reduction reaction. It is a redox reaction.

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(A) The given balanced reaction is,

$H_2SeO_4(aq)+2Cl^-(aq)+2H^+(aq)\rightarrow H_2SeO_3(aq)+Cl_2(g)+H_2O(l)$

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which chlorine shows oxidation and selenium shows reduction.

(B) The given balanced reaction is,

$S_8(s)+8O_2(g)\rightarrow 8SO_2(g)$

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which sulfur shows oxidation and oxygen shows reduction.

(C) The given balanced reaction is,

$3Br_2(aq)+6OH^-(aq)\rightarrow 5Br^-(aq)+BrO_3^-(aq)+3H_2O(l)$

This reaction is a disproportionation reaction because in this reaction only one reactant bromine that shows both oxidation and reduction reaction.

(D) The given balanced reaction is,

$Ca^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CaSO_4(s)$

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of calcium and sulfate.

(E) The given balanced reaction is,

$PtCl_4(s)+2Cl^-(aq)\rightarrow PtCl_6^{2-}(aq)$

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of platinum and chlorine.

Hence, the only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

6 0

3 years ago

Question 3 0

Tanya [424]

Answer:

Option D. KBr < KCl < NaCl

Explanation:

We'll begin by calculating the number of mole of each sample.

This can be obtained as follow:

For NaCl:

Mass = 1 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 1/58.5

Mole of NaCl = 0.0171 mole

For Kbr:

Mass = 1 g

Molar mass of KBr = 39 + 80 = 119 g/mol

Mole of KBr =?

Mole = mass /Molar mass

Mole of KBr = 1/119

Mole of KBr = 0.0084 mole

For KCl:

Mass = 1 g

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1/74.5

Mole of KCl = 0.0134 mole

Summary

Sample >>>>>>>> Number of mole

NaCl >>>>>>>>>> 0.0171

KBr >>>>>>>>>>> 0.0084

KCl >>>>>>>>>>> 0.0134

Arranging the number of mole of the sampl in increasing order, we have:

KBr < KCl < NaCl

5 0

3 years ago

To make a sarurated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 k . Find its concentration at this t

nexus9112 [7]

Answer:

The answer is " $26.47\%$ "

Explanation:

Sodium chloride solute mass $=36 \ g$

Solvent water mass $=100\ g$

Calculating the solution mass = Solute mass + solvent mass

$= 36\ g +100\ g\\\\ = 136\ g\\\\$

Calculating the percentage of concentration:

$= \frac{solute\ mass}{solvent\ mass} \times 100\\\\=\frac{36\ grm}{136\ grm} \times 100\\\\=0.2647 \times 100\\\\=26.47 \%$

3 0

3 years ago