Question

What mass of CaCO3 is required to react completely with 25.0 mL of 0.750 M HCl?

Answer 1

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So here is your answer :

moles of HCL = HCL = 0.75X 0.025 => 0.01975 malas → Cach + CO2 + H₂O Cacoz + anch 2 Males of Hal sequinel = 1 male 4 Canoz 0.01875 moles of Hel sequibus - 1x 0. 01875 2 0.009.315 mass of Cacoz = 0.009375 X 100 0.9375 , CaCO3 Thanks,

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