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drippyyahja
20.11.2020 •
Chemistry
Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3? 2HCl + CaCO3 --> CaCl2 + CO2 + H2O
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Ответ:
41 mL
Explanation:
Given data:
Milliliter of HCl required = ?
Molarity of HCl solution = 4.25 M
Mass of CaCO₃ = 8.75 g
Solution:
Chemical equation:
2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O
Number of moles of CaCO₃:
Number of moles = mass/molar mass
Number of moles = 8.75 g / 100.1 g/mol
Number of moles = 0.087 g /mol
Now we will compare the moles of CaCO₃ with HCl.
CaCO₃ : HCl
1 : 2
0.087 : 2/1×0.087 = 0.174 mol
Volume of HCl:
Molarity = number of moles / volume in L
4.25 M = 0.174 mol / volume in L
Volume in L = 0.174 mol /4.25 M
Volume in L = 0.041 L
Volume in mL:
0.041 L×1000 mL/ 1L
41 mL
Ответ: