Answer:
The answers to your questions are given below
Explanation:
F. Determination of the missing part of the equation.
_ I₂ + _Na₂S₂O₃ —> _ NaI + _ Na₂S₄O₆
The above equation can be balance as illustrated below:
I₂ + Na₂S₂O₃ —> NaI + Na₂S₄O₆
There are 2 atoms of I on the left side and 1 on the right side. It can be balance by writing 2 before NaI as shown below:
I₂ + Na₂S₂O₃ —> 2NaI + Na₂S₄O₆
There are 2 atoms of Na on the left side and a total of 4 atoms on the right side. It can be balance by writing 2 before Na₂S₂O₃ as shown below:
I₂ + 2Na₂S₂O₃ —> 2NaI + Na₂S₄O₆
Now the equation is balanced.
The coefficients are: 1, 2, 2, 1
Elements >>> Reactant >>> Product
Sodium >>>>> 4 >>>>>>>>> 4
Sulphur >>>>> 4 >>>>>>>>> 4
Oxygen >>>>> 6 >>>>>>>>> 6
Iodine >>>>>> 2 >>>>>>>>>> 2
G. Determination of the missing part of the equation.
__ Mg + __P₄ —> __Mg₃P₂
The above equation can be balance as illustrated below:
Mg + P₄ —> Mg₃P₂
There are 2 atoms of P on the right side and 4 atoms on the left side. It can be balance by writing 2 before Mg₃P₂ as shown below:
Mg + P₄ —> 2Mg₃P₂
There are 6 atoms of Mg on the right side and 1 atom on the left side. It can be balance by writing 6 before Mg as shown below:
6Mg + P₄ —> 2Mg₃P₂
Now the equation is balanced.
The coefficients are: 6, 1, 2
Elements >>>> Reactant >>> Product
Magnesium >> 6 >>>>>>>>> 6
Phosphorus >> 4 >>>>>>>>> 4
