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How many milliliters of 0.383 m hcl are needed to react with 16.2 g of caco3?

Sagot :

SOLUTION:

Step 1: Write the balanced chemical equation.

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Step 2: Calculate the volume of HCl solution needed by using dimensional analysis.

Based on the balanced chemical equation, 1 mole of CaCO₃ is stoichiometrically equivalent to 2 moles of HCl.

The molar mass of CaCO₃ is 100.086 g/mol.

[tex]\begin{aligned} \text{volume of HCl} & = \text{16.2 g} \: \text{CaCO}_3 \times \frac{\text{1 mol} \: \text{CaCO}_3}{\text{100.084 g} \: \text{CaCO}_3} \times \frac{\text{2 mol HCl}}{\text{1 mol} \: \text{CaCO}_3} \times \frac{\text{1 L soln}}{\text{0.383 mol HCl}} \times \frac{\text{1000 mL soln}}{\text{1 L soln}} \\ & = \boxed{\text{845 mL}} \end{aligned}[/tex]

Hence, 845 mL of HCl is needed to react with 16.2 g of CaCO₃.

[tex]\\[/tex]

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