finding the mass

Question

Determine the mass of excess reagent remaining after 10.0 mL of 2.50 M sodium hydroxide reacts with 23.8 mL of 1.00 M calcium chloride.

Answer 1

Hi Iris,

Step 1: First you want to determine how many moles you have of the two reagants (NaOH and CaCl₂).

You do this using the formula:

V₁C₁ = V₂C₂, where V is volume (in L), and C is concentration M in moles/L

For NaOH, 0.01L *( 2.5 mole/L) = 0.025 moles of NaOH

For CaCl₂, 0.0238L *(1.00 mole/L) = 0.0238 moles CaCl₂

Step 2: Next, write the balanced chemical reaction:

2NaOH + CaCl₂ → 2NaCl + Ca(OH)₂

According to the balanced chemical equation, you need only 1 mole of CaCl₂ for every 2 moles of NaOH.

Step 3: Next, figure out which reagant is excess from the amounts you found in Step 1 . One way to do this is with proportions:

2 moles NaOH 0.025 moles NaOH

------------------- = -------------------------

1 mole CaCl₂ X moles CaCl₂

Solve this for X = 0.0125 moles

Since X = 0.0125, then only 0.0125 moles of CaCl₂ will react with 0.025 moles NaOH.

The rest of the CaCl₂ will be excess: that's 0.0238 moles - 0.0125 moles = 0.0113 moles of CaCl₂ left over.

Step 4: Finally, convert 0.0113 moles of CaCl₂ to grams of CaCl₂ :

0.0113 moles CaCl₂ * (110.98 g/mole) = 1.254g CaCl₂

Hope that helps!

Let me know if you have any questions.

Stephanie