Balance the following equation M n ( N O 3 ) 3 ( a q ) + N a 2 S ( a q ) [Solved]

Question 1

Balance the following equation: M n ( N O 3 ) 3 ( a q ) + N a 2 S ( a q ) → M n 2 S 3 ( s ) + N a N O 3 ( a q ) Balanced Chemical Equation: A chemical equation must be balanced to satisfy the law of conversation of mass. A balanced chemical equation contains an equal number of atoms of reactants and products on both sides of the reaction.
Question 2

Is the cation in M n 2 S 3 manganese(II) or manganese(III)? Explain. Manganic sulphide: Manganic sulphide, M n 2 S 3 is an inorganic chemical compound made up of two atoms of manganese and three atoms of sulfur. The molecular weight of this compound is 206.07 g/mol. The oxidation state of manganese can be determined by knowing the common oxidation state for sulfur.
Question 3

how many moles of oxygen atoms are in 7.41 moles of Mn2(SO3)3
Question 4

Circle all the true statements below:

  1. A)  The oxidation number of S8 is 0
  2. B)  The oxidation number of Mn in Mn2S3 is -3
  3. C)  The oxidation number of S in Mn2S3 is -2
  4. D)  The oxidation number of hydrogen is always -1


Answer to question 1

The chemical equation is given below: M n ( N O 3 ) 3 ( a q ) + N a 2 S ( a q ) → M n 2 S 3 ( s ) + N a N O 3 ( a q ) Firstly, write the number of atoms for each element separately for the reactant and product sides.

The number of atoms for each element is unequal. To balance the equation, coefficients are placed in front of reactants and products to balance the equation. The coefficient 2 is placed before manganese nitrate, such that atoms of manganese got balanced, but atoms of nitrogen and oxygen increased on the reactant side. To balance the nitrogen and oxygen atoms, coefficient 6 is placed before sodium nitrate. Hence atoms of both elements are balanced. The remaining atoms of sulfur and sodium atoms are balanced by placing 3 as a coefficient in front of sodium sulfide on the side of reactants. The balanced equation is:


2Mn(NO3)3+3Na2S→Mn2S3+6NaNO32Mn(NO3)3+3Na2S→Mn2S3+6NaNO3

Now, the numbers of atoms on both sides are equal, hence the equation is balanced.


Answer to question 2

The given compound is Mn2S3Mn2S3.

Here, the cation is Mn3+Mn3+ and the anion is S2−S2−. The +3 charge on manganese comes from the anion and the -2 charge on the anion comes from the cation. The oxidation state of anion must be written on the subscript of the cation and the oxidation state of cation must be written at the subscript of anion to make a neutral molecule. The mathematical formulation is shown below.

On manganese the charge is +3. Therefore, this number must be written in Roman numeral. Thus, the cation in Mn2S3Mn2S3 is manganese (III).

Hence, the cation in Mn2S3Mn2S3 is manganese (III).


Answer to question 3

Step 1

The molecular formula of the given compound is = Mn2(SO3)3

The molecular formula of the compound states that one molecule of manganese sulfide contains = 3×3 = 9 oxygen atoms

Step 2

Mole concept:

The number of molecules or atoms present in the one mole of the substance is equal to the Avogadro number. The value of the Avogadro number is 6.022×1023. The mass of one mole of a substance is known as the molar mass.

Step 3

Before we calculate the number of oxygen atoms, first we calculate the number of molecules of Mn2(SO3)3.

We know that that one mole of Mn2(SO3)3 contains = 6.022×1023 molecules of Mn2(SO3)3

One mole of Mn2(SO3)3 contains = 6.022×1023 molecules of Mn2(SO3)37.41 moles of Mn2(SO3)3 contains = 6.022×1023×7.41 molecules of Mn2(SO3)37.41 moles of Mn2(SO3)3 contains = 44.62302×1023 molecules of Mn2(SO3)3

Step 4

We know that one molecule of Mn2(SO3)3 contais nine oxygen atoms.

Hence the number of oxygen atom is calculated as follows:

One molecule of Mn2(SO3)3 contains = 9 molecules of oxygen 44.62302×1023 molecules of Mn2(SO3)3  contains = 9×44.62302×1023 molecules of oxygen 44.62302×1023 molecules of Mn2(SO3)3  contains = 401.60718×1023 molecules of oxygen 44.62302×1023 molecules of Mn2(SO3)3  contains = 4.02×1025 molecules of oxygen

Final answer

The number of the oxygen atom in 7.41 moles of Mn2(SO3)is =  4.02×1025 molecules of oxygen


Answer to question 4

Step 1

Inorganic chemistry.

Step 2

1. A) The oxidation number of S8 is 0 – True

2. B) The oxidation number of Mn in Mn2S3 is -3 – False; The oxidation number of Mn in Mn2S3 is +3

3. C) The oxidation number of S in Mn2S3 is -2 – True.

4. D) The oxidation number of hydrogen is always -1 – False; The oxidation number of hydrogen is mostly +1 except in few cases like metal hydrides (eg. NaH).

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