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What mass of hydrochloric acid (HCI) is needed to react with 8.2 g Caco,?

__CaCO3 + HCl →_CaCl₂ + H₂O+_CO₂

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Answer:

The balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂.

The molar mass of CaCO₃ is 100.0869 g/mol 1. Therefore, 8.2 g of CaCO₃ is equivalent to 0.0819 moles of CaCO₃.

Since the reaction requires 2 moles of HCl for every mole of CaCO₃, 0.0819 moles of CaCO₃ will require 0.1638 moles of HCl.

The molar mass of HCl is 36.46094 g/mol 2. Therefore, 0.1638 moles of HCl is equivalent to 5.97 g of HCl.

So, you will need approximately 5.97 g of hydrochloric acid (HCl) to react with 8.2 g of calcium carbonate (CaCO₃).

Step-by-step explanation:

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