Moose Concentration oF NaOH = 1.545 Table #2 Trial #1 Trial #2 Trial #3 Total Vo

Question

Moose Concentration oF NaOH = 1.545 Table #2 Trial #1 Trial #2 Trial #3 Total Volume of NaOH Added 36.90 32.50 32.00 % Calcium Carbonate per tablet Mass of carbonate per tablet Sample # and Mass (g) Sample 1 1.304 Sample 2 - 1.275 Sample 3 -1.327 Show a balanced equation for the following chemical reaction .What gas is formed from when you drop Tums into HCl? How many moles of that gas should be produced if 500 mg of carbonate is assumed? HCl Concentration 0.4915 Calculate mass and % mass of calcium carbonate in each tablet. Perform separate calculations for each sample individually. Show calculations for each trial

Explanation / Answer

Q1

a)

gas of HCl + tumbs --> CO2(g) from the CaCO3 + 2H+ = H2O + Ca+2 + CO2 interaction

the chemical formula

CaCO3(s) + 2HCl(aq) = H2O(l) + CaCl2(aq) + CO2(g)

b)

moles of gas if

500 mg of carbonate are consumed

mol of Carbonate ( CO3-2) = mass/M W= 0.5/56 = 0.0089285

then, 1 mol of CO3-2 = 1 mol of CO2

0.0089285 mol of CO3-2 = 0.0089285 mol of CO2

c)

% mass of each CaCO3 tablet

trial 1 mol of HCl = MV = 0.4915*(36.90*10^-3) = 0.018136

trial 2 mol of HCl = MV = 0.4915*(32.50*10^-3) = 0.01597

trial 3 mol of HCl = MV = 0.4915*(32*10^-3) = 0.015728

now...

2 mol of HCl = 1 mol of CaCO3

mol of CaCO3 :

trial 1 mol of CaCO3 = 1/2*mol of HCl = 0.018136*1/2 = 0.009068

trial 1 mol of CaCO3 = 1/2*mol of HCl = 0.01597*1/2 = 0.007985

trial 1 mol of CaCO3 = 1/2*mol of HCl = 0.015728*1/2 = 0.007864

mass of CacO3:

mass of CaCO3 = 0.009068*100 = 0.9068 g

mass of CaCO3 =0.007985 *100 = 0.7985 g

mass of CaCO3 =0.007864*100 = 0.7864 g

% mass:

trial 1 --> CaCO3% = mass/Total mass * 100% = 0.9068 /1.304*100 = 69.5 %

trial 2 --> CaCO3% = mass/Total mass * 100% = 0.7985 /1.275*100 = 62.6 %

trial 3 --> CaCO3% = mass/Total mass * 100% = 0.7864 /1.327*100 = 59.6 %

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