What volume of 0.80 M HCl would be needed to react with 4.0 g of CaCO3?

1 Answer
Andrea B.
Apr 19, 2018

62.5 mL

Explanation:

The reaction is
#2HCl + CaCO_3 = CaCl_2 + CO_2 + H_2O#
if you have 4 g of #CaCO_3 # you have #(4g)/(100g/(mol))= 0.025 mol# of salt and for the stechiometry of the reaction you must have the double of HCl that is 0.050 mol
As for a solution #M= n/V# so you have #V= n/M=(0.05 mol)/ (0.80 (mol)/(L))= 0,0625 L =62,5 mL#

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