Question

Limiting Reactant Calculations Reaction #3 CaCl2: 0.250m, 5.00mL KOH: 0.350M 4.99 mL CaCl2 + KOH → KCI + Ca(OH)2 How many moles of CaCl are in the initial reaction? Show your work. How many moles of ions would that be? moles CaCl2 moles Ca2+ moles Cr How many moles of KOH are in the initial reaction? Show your work. How many moles of ions would that be? moles KOH moles K moles OH What is the limiting reactant? How can you tell based on your observations? Using the table provided in appendix H, look up the solubility values for the two products KCI: Ca(OH)2: Based on these numbers, what would the solid product be? What would be the theoretical yield of the reaction? Show your work. Using the theoretical yield and the solubility value for the precipitate, how many mg of precipitate would be in your test tube? Show your work,

Answer 1

To determine the limiting reactant in this reaction, compare the number of moles of CaCl2 and KOH. CaCl2 is the limiting reactant because it produces fewer moles compared to KOH.

Step-by-step explanation:

In order to determine the limiting reactant in this reaction, we need to compare the number of moles of CaCl2 and KOH. Let's start with CaCl2. Given that the concentration is 0.250 M and the volume is 5.00 mL, we can calculate the number of moles using the formula:

moles = concentration x volume

moles of CaCl2 = 0.250 M x 0.00500 L = 0.00125 moles

Next, let's calculate the number of moles of KOH. Given that the concentration is 0.350 M and the volume is 4.99 mL, we can use the same formula:

moles of KOH = 0.350 M x 0.00499 L = 0.00174 moles

Based on these calculations, we can see that CaCl2 is the limiting reactant because it produces fewer moles compared to KOH.

Answer 2

The initial reaction contains 0.0125 moles of CaCl2. The moles of ions for CaCl2 are 0.025 moles Ca2+ and 0.025 moles Cl-. The initial reaction has 0.0175 moles of KOH, resulting in 0.0175 moles K+ and 0.0175 moles OH-. The limiting reactant is CaCl2, evident from the balanced equation stoichiometry. Using solubility values, KCI is soluble, while Ca(OH)2 is slightly soluble. The theoretical yield is determined as per the limiting reactant, and based on solubility, Ca(OH)2 is the solid product. Theoretical yield calculation yields 0.0186 moles of Ca(OH)2, equivalent to 0.373 g. Following solubility, the precipitate in the test tube would be 36.9 mg.

Step-by-step explanation:

To find the moles of CaCl2, multiply its concentration (0.250 M) by the volume (5.00 mL) and convert to moles. Repeat for KOH. Calculate moles of ions for each compound by multiplying the moles of the compound by the number of ions it yields. Identify the limiting reactant by comparing the moles of reactants to the stoichiometric coefficients.

Check solubility values in Appendix H to determine the solid product. The theoretical yield is found by multiplying the moles of the solid product by its molar mass. To find the mass of precipitate, multiply the theoretical yield by 1000 (for grams to milligrams).