How many mols $CaCO_3$ can be dissolved in .0250 mol $HCl$ in the equation $CaCO_3 + 2HCl -> CaCl_2 + H_2O + CO_2$ ?

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Answer 1 · 2016-04-09T13:42:27

$0.0125molCaCO_3$

The equation says

$CaCO_3 + 2HCl$ ,

which means that for every $2$ moles of $HCl$ there is one of $CaCO_3$ . This is called the molar ratio, in this case $2:1$ .

We know that there are $0.0250$ moles of $HCl$ , which is equal to two parts in the ratio. If $0.0250$ is two parts, then one part must be $0.0125$ .

And, since there is only one part of $CaCO_3$ in the ratio, then you can dissolve $0.0125$ moles of $CaCO_3$ in $0.0250$ moles of $HCl$ .

How many mols CaCO_3CaCO_3 can be dissolved in .0250 mol HClHCl in the equation CaCO_3 + 2HCl -> CaCl_2 + H_2O + CO_2CaCO_3 + 2HCl -> CaCl_2 + H_2O + CO_2?