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Calcium carbonate reacts with aqueous HCl according to the reaction : CaCO_(3)(s)+2HCl(aq) rarr CaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l). What mass of CaCO_(3) is required to react completely with 25 mL of 0.75 M HCl ? |
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Answer» SOLUTION :Step 1. To calculate mass of HCl in 25 mL of 0.75 M Hcl 1000 mL of 0.75 M HCl contains HCl = 0.75 moles `=0.75xx36.5g=24.375g` `therefore"25 mL of 0.75 HCl will contain HCl"=(24.375)/(1000)xx25g=0.6844g.` Step 2. To calculate mass of `CaCO_(3)` reacting completely with 0.9125 G of HCl `CaCO_(3)(s)+2HCl(aq) rarr CaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l)` 2 moles of HCl, i.e., `2xx36.5 g=73g HCl` REACT completely with `CaCO_(3) = "1 MOLE = 100 g"` `therefore " 0.6844 g HCl will react completely with "CaCO_(3)=(100)/(73)xx0.6844g=0.938g` |
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