Answer:
The total cost is 1.25 dollars.
Step-by-step explanation:
The reaction between HCl and CaCO₃ is giving by:
2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + CO₂(g) + H₂O(l) (1)
0.500L M: 100.01g/mol
0.400M
According to equation (1), 2 moles of HCl react with 1 mol of CaCO₃, so to neutralize HCl, we need the next amount of CaCO₃:
[tex] m CaCO_{3} = (\frac{1 \cdot mol HCl}{2}) \cdot M_{CaCO_{3}} = (\frac{0.500L \cdot 0.400 \frac {mol}{L}}{2}) \cdot 100.01 \frac{g}{mol} = 10.001 g [/tex]
The CaCO₃ mass of each tablet is:
[tex] m CaCO_{3} = 1 g_{tablet} \cdot \frac{40g CaCO_{3}}{100g_{tablet}} = 0.4g [/tex]
Hence, the number of tablets that we need to neutralize the HCl is:
[tex] number_{tablets} = ( \frac{1 tablet}{0.4 g CaCO_{3}}) \cdot 10.001g CaCO_{3} = 25 [/tex]
Finally, if every 80 tablets costs 4.00 dollars, 25 tablets will cost:
[tex] cost = (\frac {4 dollars}{80 tablets}) \cdot 25 tablets = 1.25 dollars [/tex]
So, the total cost to neutralize the HCl is 1.25 dollars.
I hope it helps you!
