# 1.1 Introduction to the particulate nature of matter and chemical change SL Paper 1

### Question

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

__Na2S2O3(aq) + __HCl(aq) → __S(s) + __SO2(g) + __NaCl(aq) + __H2O(l)

A.     6

B.     7

C.     8

D.     9

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.

Na2S2O3(aq) + HCl(aq) → S(s) + SO2(g) + 2NaCl(aq) + H2O(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 :

1Na2S2O3(aq) + 2HCl(aq) → 1S(s) + 1SO2(g) + 2NaCl(aq) + 1H2O(l)

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

### Question

Which is a homogeneous mixture?

A.     Oil and water

B.     Sand and water

C.     Ethanol and water

D.     Chalk and sand

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.

### Question

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

__C6H12O6 (aq) → __C2H5OH (aq) + __CO2 (g)

A. 4

B. 5

C. 9

D. 10

B

Number of H atoms in reactants side is 12 and in products side is 6. Hence, place the coefficient 2 in front of C2H5OH.

C6H12O6 (aq) → 2C2H5OH (aq) + CO2 (g)

Now,  There are 6 C atoms in reactants and 5 in products. Place 2 in front of CO2.

C6H12O6 (aq) → 2C2H5OH (aq) + 2CO2 (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?

C8H18(g) + O2(g) → CO(g) + H2O(l)

A.     26.5

B.     30

C.     53

D.     61

C

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

C8H18(g) + O2(g) →  8CO(g) + H2O(l)

Number of H atoms in reactants side is 18 and in products side is 2. Hence, place the coefficient 9 in front of H2O.

C8H18(g) + O2(g) →  8CO(g) + 9H2O(l)

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

Place 8.5 in front of O2.

C8H18(g) + 8.5O2(g) →  8CO(g) + 9H2O(l)

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

2C8H18(g) + 17O2(g) →  16CO(g) + 18H2O(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?

__C6H12O6 (aq) → __C2H5OH (aq) + __CO2 (g)

A. 4

B. 5

C. 9

D. 10

B

Number of H atoms in reactants side is 12 and in products side is 6. Hence, place the coefficient 2 in front of C2H5OH.

C6H12O6 (aq) → 2C2H5OH (aq) + CO2 (g)

Now,  There are 6 C atoms in reactants and 5 in products. Place 2 in front of CO2.

C6H12O6 (aq) → 2C2H5OH (aq) + 2CO2 (g)

Now the equation is balanced.

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

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