Question

if the below reaction occurs and the equation is properly balanced with the smallest whole - number coefficients, what is coefficients for br?
al(s) + br₂(l) → al₂br₆(s)
add your answer
integer, decimal, or e notation allowed
question 2
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?
b₂o₃(s) + hf(l) → bf₃(g) + h₂o(l)
add your answer
integer, decimal, or e notation allowed
question 3
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)
add your answer
integer, decimal, or e notation allowed

Explanation:

Question 1

Step1: Balance the equation for \(Al(s)+Br_2(l)\to Al_2Br_6(s)\)

• Balance \(Al\) atoms: There are \(2\) \(Al\) atoms on the product side. So, put a coefficient of \(2\) in front of \(Al\) on the reactant side: \(2Al(s)+Br_2(l)\to Al_2Br_6(s)\)
• Balance \(Br\) atoms: There are \(6\) \(Br\) atoms on the product side. Since each \(Br_2\) molecule has \(2\) \(Br\) atoms, put a coefficient of \(3\) in front of \(Br_2\): \(2Al(s)+3Br_2(l) = Al_2Br_6(s)\)

Question 2

Step1: Balance the equation for \(B_2O_3(s)+HF(l)\to BF_3(g)+H_2O(l)\)

• Balance \(B\) atoms: There are \(2\) \(B\) atoms on the reactant side. Put a coefficient of \(2\) in front of \(BF_3\): \(B_2O_3(s)+HF(l)\to 2BF_3(g)+H_2O(l)\)
• Balance \(F\) atoms: There are \(6\) \(F\) atoms on the product side. Put a coefficient of \(6\) in front of \(HF\): \(B_2O_3(s)+6HF(l)\to 2BF_3(g)+H_2O(l)\)
• Balance \(O\) and \(H\) atoms: There are \(3\) \(O\) atoms on the reactant side. Put a coefficient of \(3\) in front of \(H_2O\). The balanced equation is \(B_2O_3(s)+6HF(l)=2BF_3(g)+3H_2O(l)\)
• Sum of coefficients: \(1 + 6+2 + 3=12\)

Question 3

Step1: Balance the equation for \(C_6H_6(l)+O_2(g)\to H_2O(g)+CO_2(g)\)

• Balance \(C\) atoms: There are \(6\) \(C\) atoms in \(C_6H_6\). Put a coefficient of \(6\) in front of \(CO_2\): \(C_6H_6(l)+O_2(g)\to H_2O(g)+6CO_2(g)\)
• Balance \(H\) atoms: There are \(6\) \(H\) atoms in \(C_6H_6\). Put a coefficient of \(3\) in front of \(H_2O\): \(C_6H_6(l)+O_2(g)\to 3H_2O(g)+6CO_2(g)\)
• Balance \(O\) atoms: There are \(3 + 12=15\) \(O\) atoms on the product side. Since \(O_2\) has \(2\) \(O\) atoms, multiply the entire equation by \(2\) to get rid of the fraction. The balanced equation is \(2C_6H_6(l)+15O_2(g)=6H_2O(g)+12CO_2(g)\)
• Sum of coefficients: \(2+15 + 6+12=35\)

Answer:

Question 1: \(3\)
Question 2: \(12\)
Question 3: \(35\)