3. Based on question 2 (above), if the BaCl2 solution used was 0.11 M, how many

Question

3. Based on question 2 (above), if the BaCl2 solution used was 0.11 M, how many moles of BaCl2 will be left over after the reaction has completed? (4 pts) 4. What indication in the "test tube test" (procedural step 9) will be given if Na3PO4 is the limiting reagent? What if BaCl2 is the limiting reagent? Explain why. (3 pts) Clearly demonstrate which reactant salt would be limiting if the experiment was performed with solutions of equal concentrations. Assume 20 mL each of 0.5 M solutions, for instance Show calculations. (4 pts)

Explanation / Answer

Ans 2

Moles of BaCl2 = molarity x volume

= 0.11 mol/L x 0.035 L

= 0.00385 mol

2 Na3PO4 + 3 BaCl2 = Ba3(PO4)2 + 6 NaCl

Mol of Ba3(PO4)2 formed = mass/molecular weight

= 0.520 g / 601.93 g/mol

= 0.00086388 mol

Moles of Na3PO4 consumed = 2*0.00086388

= 0.0017277 mol

Moles of BaCl2 consumed = 0.0017277 x 3/2 = 0.00259 mol

Moles of BaCl2 left over = 0.00385 - 0.00259

= 0.00126 mol

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