Question

0.700 g of an impure fecl₃ sample are dissolved in water to create 50.0 ml of solution. the sample was titrated with 15.41 ml of 0.100 m h₂c₂o₄ to reach the equivalence point.
how many moles of h₂c₂o₄ react to reach the equivalence point?
? × 10^? mol h₂c₂o₄

Explanation:

Step1: Recall the formula for moles from molarity and volume

The formula to calculate moles (\(n\)) from molarity (\(M\)) and volume (\(V\)) in liters is \(n = M\times V\).

Step2: Convert volume to liters

The volume of \(\ce{H2C2O4}\) is \(15.41\space mL\). Since \(1\space L = 1000\space mL\), we convert it to liters: \(V=\frac{15.41\space mL}{1000\space mL/L}=0.01541\space L\).

Step3: Calculate moles of \(\ce{H2C2O4}\)

Given \(M = 0.100\space M\) (or \(0.100\space mol/L\)) and \(V = 0.01541\space L\), we use the formula \(n = M\times V\).
\(n=0.100\space mol/L\times0.01541\space L = 0.001541\space mol\).
We can write this in scientific notation as \(1.541\times 10^{- 3}\space mol\).

Answer:

\(1.541\times 10^{-3}\) (so the first box is \(1.541\) and the second box is \(-3\))