Question

if the below reaction occurs and the equation is properly balanced with the smallest whole - number coefficients, what is the sum of the coefficients for all reactants and products?
c₆h₆(l) + o₂(g) → h₂o(g) + co₂(g)
35
integer, decimal, or e notation allowed
question 4
if the below reaction occurs and the equation is properly balanced with the smallest whole - number coefficients, what is the sum of the coefficients for all reactants and products?
uo₂(s) + hf(l) → uf₄(s) + h₂o(l)
add your answer
integer, decimal, or e notation allowed

Explanation:

Step1: Balance the chemical equation

For the reaction \(UO_{2}(s)+HF(l)\to UF_{4}(s)+H_{2}O(l)\), balance the atoms.

• Balance \(F\) atoms: There are \(4\) \(F\) atoms in \(UF_{4}\), so put a coefficient of \(4\) in front of \(HF\). The equation becomes \(UO_{2}(s)+4HF(l)\to UF_{4}(s)+H_{2}O(l)\).
• Balance \(H\) and \(O\) atoms: There are \(4\) \(H\) atoms from \(HF\). On the right - hand side, in \(H_{2}O\), if we put a coefficient of \(2\) in front of \(H_{2}O\), we have \(4\) \(H\) atoms (\(2\times2\)) and \(2\) \(O\) atoms (which matches the \(2\) \(O\) atoms in \(UO_{2}\)). The balanced equation is \(UO_{2}(s)+4HF(l)=UF_{4}(s)+2H_{2}O(l)\).

Step2: Calculate the sum of coefficients

The coefficients of the reactants and products are \(1\) (\(UO_{2}\)), \(4\) (\(HF\)), \(1\) (\(UF_{4}\)), and \(2\) (\(H_{2}O\)).
The sum of the coefficients is \(1 + 4+1 + 2\).

$$1+4 + 1+2=\sum_{i = 1}^{4}a_{i}$$

where \(a_{1}=1\), \(a_{2}=4\), \(a_{3}=1\), \(a_{4}=2\)

$$1+4+1 + 2=8$$

Answer:

\(8\)