Question

write the formula for zinc acetate. *
your answer
how many moles of febr₃ can be produced from 4.32 moles of fe? *
unbalanced equation
fe + br₂ → febr₃
your answer
write the formula for hydrobromic acid. *
your answer
name the compound h₂so₃. *
your answer

Explanation:

Step1: Write the formula for zinc acetate

Zinc has a charge of \(+2\) (\(Zn^{2+}\)) and acetate ion (\(C_{2}H_{3}O_{2}^{-}\)) has a charge of \(-1\). To balance the charges, we need 2 acetate ions for each zinc ion. So the formula is \(Zn(C_{2}H_{3}O_{2})_{2}\).

Step2: Calculate moles of \(FeBr_{3}\) from moles of \(Fe\)

First, balance the equation: \(2Fe + 3Br_{2}
ightarrow2FeBr_{3}\). From the balanced equation, the mole ratio of \(Fe\) to \(FeBr_{3}\) is \(1:1\). Given \(4.32\) moles of \(Fe\), using the mole - ratio \(\frac{n(FeBr_{3})}{n(Fe)}=\frac{1}{1}\), so \(n(FeBr_{3}) = 4.32\) moles.

Step3: Write the formula for hydrobromic acid

Hydrobromic acid is a binary acid. The formula for binary acids is \(H\) followed by the non - metal. Bromine has a charge of \(-1\) and hydrogen has a charge of \(+1\). So the formula is \(HBr\).

Step4: Name the compound \(H_{2}SO_{3}\)

\(H_{2}SO_{3}\) is an oxy - acid. The anion \(SO_{3}^{2 -}\) is sulfite. The naming rule for oxy - acids: if the anion ends with “-ite”, the acid ends with “-ous”. So the name is sulfurous acid.

Answer:

Zinc acetate: \(Zn(C_{2}H_{3}O_{2})_{2}\)
Moles of \(FeBr_{3}\): \(4.32\)
Hydrobromic acid: \(HBr\)
\(H_{2}SO_{3}\): Sulfurous acid