Write the balanced net ionic equation for the reaction of HCI with the active in
ID: 866710 • Letter: W
Question
Write the balanced net ionic equation for the reaction of HCI with the active ingredient in your tablet Active Ingredient(s): Na2 Co3+2HCl rightarrow 2Nacl+Co2+H2o Calculate the cost of neutralization in dollars per mole of acid neutralized by one (cost per tablet: $7.28 for 1000 tablets Answer: 0.375 Using the mass of active ingredient (antacid) per tablet found on the antacid container, calculate the volume of 0.10 M HCI that can be neutralized by one tablet, Assuming that the density of 0.10 M HCI is 1.0 g/mL, what mass of 0.10 M acid can be neutralized by one tablet? (c) Determine the ratio of the mass of the HCI(aq) solution to the mass of one tablet. (average of your 2 tabs) Volume 0.10 M HCI MgCaCo3 rightarrow gCaCo3 rightarrow MolCaCo3 rightarrow MolCaCo3 rightarrow MolHCI rightarrow rightarrow CHCI Mass 0.10 M HCI , 10 Times 34.5 = 3.45 HCl rightarrow ML HCL rightarrow gHCl Number of times its weight 5 (show calc's below) 2moles of HCL rightarrow moleCaCo3 73gm of HCL required for 100gm of CaCo3 CaCo3 required in milligram is 500gm Calculate the mass of ACTIVE INGREDIENT per tablet based on your moles of H+ neutralized per tablet. Compare with the information given on the label. (Formula of ACTIVE INGREDIENT: =2.59g/tablet mass active ingredient Answer:Explanation / Answer
if the mass of Na2CO3 is 0.375g in one tablet you can calculate the number of mol:
n = m/M = 0.375g/106g/mol = 3.54*10-3 mol per tablet
you see from the reaction that: n(Na2CO3) = 2n (HCl)
n(HCl) = 3.54*10-3 mol/2 = 1.77*10-3 mol (the number of mol which can be neutralized by one tablet)
c= n/V to calculate the Volume: V= n/c= 1.77*10-3 mol/ 0.1mol/L = 0.0177 L = 17.7mL
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