Question

pre - lab questions
answer the following questions before attending lab. show your work fully and clearly.

1. define stoichiometry.
2. calculate the number of moles of hcl in 40.0 ml of 1.0 m hcl and the number of moles of naoh in 15.0 ml of 1.0 m naoh.
3. study the dilution calculations on pages 325 - 328 of your textbook. then do the necessary calculations and explain how to make the following solutions. a complete answer requires a procedure in addition to a number.

a. 250 ml of 1.0 m hcl using 5.0 m hcl
b. 250 ml of 1.0 m h₂so₄ using 6.0 m h₂so₄

Explanation:

Step1: Recall the formula for moles

The formula for moles \(n\) is \(n = M\times V\), where \(M\) is the molarity and \(V\) is the volume in liters.

Step2: Convert volumes to liters

For \(HCl\): \(V = 40.0\space mL=40.0\times10^{- 3}\space L = 0.0400\space L\). For \(NaOH\): \(V = 15.0\space mL = 15.0\times10^{-3}\space L=0.0150\space L\)

Step3: Calculate moles of \(HCl\)

Using \(n = M\times V\), with \(M = 1.0\space M\) and \(V = 0.0400\space L\), we have \(n_{HCl}=1.0\space M\times0.0400\space L = 0.040\space mol\)

Step4: Calculate moles of \(NaOH\)

Using \(n = M\times V\), with \(M = 1.0\space M\) and \(V = 0.0150\space L\), we have \(n_{NaOH}=1.0\space M\times0.0150\space L=0.015\space mol\)

Answer:

The number of moles of \(HCl\) is \(0.040\space mol\) and the number of moles of \(NaOH\) is \(0.015\space mol\)