*Question*

What is the sum of the coefficients when the equation is balanced with the lowest whole number ratio?

__Na_{2}S_{2}O_{3}(aq) + __HCl(aq) → __S(s) + __SO_{2}(g) + __NaCl(aq) + __H_{2}O(l)

A. 6

B. 7

C. 8

D. 9

**▶️Answer/Explanation**

C

**Explanation: **Firstly, count the number of Na atoms in reactants and products side. There are 2 Na atoms in the reactants side, then there must be 2 Na atoms in the products side also. Hence, place the coefficient 2 in front of NaCl to balance Na.

Na_{2}S_{2}O_{3}(aq) + HCl(aq) → S(s) + SO_{2}(g) + 2NaCl(aq) + H_{2}O(l)

Similarly for Sulfur atoms, the number of S atoms on both sides is equal i.e. 2.

For Oxygen atoms, there are 3 O atoms on both sides.

There is only 1 H atom in the reactants, but there are two H atoms in products. Hence, place the coefficient 2 in front of HCl. After doing this, Cl atoms also get balanced on both sides.

So, the equation becomes :

1Na_{2}S_{2}O_{3}(aq) + 2HCl(aq) → 1S(s) + 1SO_{2}(g) + 2NaCl(aq) + 1H_{2}O(l)

Now, Sum of the coefficients: 1 + 2 + 1 + 1 + 2 + 1 i.e 8.

Hence, correct answer is C.

*Question*

Which is a homogeneous mixture?

A. Oil and water

B. Sand and water

C. Ethanol and water

D. Chalk and sand

**▶️Answer/Explanation**

C

Option A : Oil and water do not mix. They forms two phases as there is a boundary of separation between oil and water, it is a heterogeneous mixture.

Option B: Sand doesn’t dissolve in water, it settles down in it, hence it is a heterogeneous mixture.

Option C: Ethanol combines with water to form a single phase only and there is no boundary of separation between ethanol and water. Hence the mixture is homogeneous in nature.

Option D: Chalk doesn’t mix with sand. it forms a heterogeneous mixture.

Hence correct answer is C.

*Question*

What is the sum of the coefficients when the following equation is balanced using the smallest whole numbers?

__C_{6}H_{12}O_{6} (aq) → __C_{2}H_{5}OH (aq) + __CO_{2} (g)

A. 4

B. 5

C. 9

D. 10

**▶️Answer/Explanation**

B

Number of H atoms in reactants side is 12 and in products side is 6. Hence, place the coefficient 2 in front of C_{2}H_{5}OH.

C_{6}H_{12}O_{6} (aq) → 2C_{2}H_{5}OH (aq) + CO_{2} (g)

Now, There are 6 C atoms in reactants and 5 in products. Place 2 in front of CO_{2.}

C_{6}H_{12}O_{6} (aq) → 2C_{2}H_{5}OH (aq) + 2CO_{2} (g)

Now the equation is balanced.

Sum of the coefficients is: 1 + 2 + 2 = 5.

*Question*

What is the sum of the coefficients when the equation is balanced with whole numbers?

_{—}C_{8}H_{18}(g) + _{—}O_{2}(g) → _{—}CO(g) + _{—}H_{2}O(l)

A. 26.5

B. 30

C. 53

D. 61

**▶️Answer/Explanation**

C

There are 8 C atoms in reactants and 1 in products. Place 8 in front of CO.

C_{8}H_{18}(g) + O_{2}(g) → 8CO(g) + H_{2}O(l)

Number of H atoms in reactants side is 18 and in products side is 2. Hence, place the coefficient 9 in front of H_{2}O.

C_{8}H_{18}(g) + O_{2}(g) → 8CO(g) + 9H_{2}O(l)

Number of O atoms in reactants side is 2 and in products side is 17.

Place 8.5 in front of O_{2. }

C_{8}H_{18}(g) + 8.5O_{2}(g) → 8CO(g) + 9H_{2}O(l)

Since equation is to be balanced with whole numbers, multiply both sides by 2.

2C_{8}H_{18}(g) + 17O_{2}(g) → 16CO(g) + 18H_{2}O(l)

Now, sum of coefficients : 2 + 17 + 16 + 18 = 53.

*Question*

What is the sum of the coefficients when the following equation is balanced using the smallest whole numbers?

__C_{6}H_{12}O_{6} (aq) → __C_{2}H_{5}OH (aq) + __CO_{2} (g)

A. 4

B. 5

C. 9

D. 10

**▶️Answer/Explanation**

B

Number of H atoms in reactants side is 12 and in products side is 6. Hence, place the coefficient 2 in front of C_{2}H_{5}OH.

C_{6}H_{12}O_{6} (aq) → 2C_{2}H_{5}OH (aq) + CO_{2} (g)

Now, There are 6 C atoms in reactants and 5 in products. Place 2 in front of CO_{2.}

C_{6}H_{12}O_{6} (aq) → 2C_{2}H_{5}OH (aq) + 2CO_{2} (g)

Now the equation is balanced.

Sum of the coefficients is: 1 + 2 + 2 = 5.