Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: SO42−(aq)+Sn2+(aq)+ –––→H2SO3(aq)+Sn4+(aq)+ ––– Part A What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms.

The balanced equation is given below:

SO4^2−(aq)+ Sn^2+(aq) + 4H^+(aq) → H2SO3(aq)+ Sn^4+(aq) + H2O(l)

The coefficients of the reactants are: 1, 1, 4

The coefficients of the products are: 1, 1, 1

Explanation:

Step 1:

The equation for the reaction. This is given below:

SO4^2−(aq)+ Sn^2+(aq) + –––→ H2SO3(aq)+ Sn^4+(aq) + –––

Step 2:

Inclusion of H+ and H2O into the equation. This is illustrated below:

SO4^2−(aq)+ Sn^2+(aq) + H^+(aq) → H2SO3(aq)+ Sn^4+(aq) + H2O(l)

Step 3:

Balancing the equation.

SO4^2−(aq)+ Sn^2+(aq) + H^+(aq) → H2SO3(aq)+ Sn^4+(aq) + H2O(l)

The above equation can be balanced as follow:

There are total 4 atoms of H on the right side and 1 atom on the left side. It can be balance by putting 4 in front of H^+ as shown below:

SO4^2−(aq)+ Sn^2+(aq) + 4H^+(aq) → H2SO3(aq)+ Sn^4+(aq) + H2O(l)

Now the equation is balanced.

The coefficients of the reactants are: 1, 1, 4

The coefficients of the products are: 1, 1, 1

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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