A 0.308 g sample of Tums (containing CO 2 − 3 ) is dissolved in 25.0...
A 0.308 g sample of Tums (containing CO 2 − 3 ) is dissolved in 25.0 mL of 0.102 M HCl. The hydrochloric acid that is not neutralized by the Tums is back titrated with 8.14 mL of 0.102 M NaOH. Calculate the neutralizing capacity (in mol acid / g antacid) of Tums according to these results.
Homework Answers
The reaction between HCl and NaOH is 1:1molar
HCl + NaOH -------> NaCl + H2O
moles of NaOH consumed = (0.102mol/1000ml) × 8.14ml = 0.00083028mol
excess moles of HCl = 0.00083028mol
Total moles of HCl = (0.102mol/1000ml)× 25.0ml = 0.00255mol
moles of HCl neutralized = 0.00255mol - 0.00083028mol = 0.001720mol
Therefore,
neutralizing capacity of Tums = 0.001720mol/0.308g
= 0.005584mol acid/g antacid
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A 0.308 g sample of Tums (containing CO 2 − 3 ) is dissolved in
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this is my prelab for class, I'm really struggling to understand so if someone could help...
this is my prelab for class, I'm really struggling to understand so
if someone could help me break down the problems that would be
nice
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help on number 3 here is lab we did
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A Tums tablet weighed 1.319 g. Using the procedure of this experiment, the tablet was dissolved in, and reacted with, an excess amount of HCl (125.00 mL of 0.101 M HCl). The excess unreacted HCl was titrated with NaOH; 18.75 mL of 0.101 M NaOH was used to reach the endpoint. Calculate the moles of HCl that did not react with the antacid. Calculate the moles of HCl that did react with the antacid. Calculate the mass of HCl that...
this
is my prelab for class, I'm really struggling to understand so if
someone could help me break down the problems that would be
nice
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this is my prelab for class, I'm really struggling to understand so
if someone could help me break down the problems that would be
nice
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