Question

1. Balance the following chemical equations a + C4H10O + O2 ...CO₂ + H₂O (b) Febr3 + H2SO4 Fe2(SO4)3 + HBO (c) P4010 + H2O -- H3PO4 (d) KClO3 ---KCIO4 + KCI
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Answer 1

2C4H10(g)+13O2→8CO2(g)+10H2O(g) 

Explanation :

We have the unbalanced equation 

C4H10(g)+O2(g)→CO2(g)+H2O(g) 

This is the combustion of butane C4H10 

Let's first balance the carbons. There are four on the LHS, but only one on the RHS, so we multiply CO2 by 4 to get 

Final step is to balance the oxygens. There are two on the left hand side, but thirteen on the right hand side, so we need to divide thirteen by two to get the "scale number", which is 6.5. The equation is thus:

C4H10(g)+6.5O2g)→4CO2(g)+5H2O(g) 

But wait, we cannot have half a molecule! So, we need to multiply the whole equation by 2, which leads us to the finalized, balanced equation: