What volume of H2 will be released (at 298 K and 1 atm) if 4.008 g of Ca reacts with excess HCl?

First,barite the balanced chemical equation of the reaction: $$ Ca+2HCl\xrightarrow{}CaCl_2+H_2 $$ From the reaction equation, it is clear that from 1 mole of Ca we will get 1 mole of hydrogen gas. $$ 1\:mole\:Ca\xrightarrow{}1\:mole\:H_2 $$ The molar mass of Ca is 40 grams per mole(approx.) And of hydrogen is 2 grams per mole. So, $$ 40\:g\:Ca\xrightarrow{}2\:g\:H_2 $$ $$ 1\:g\:Ca\:\xrightarrow{}\frac{2}{40}\:g\:H_2=\frac{1}{20}g\:H_2 $$ Also, we have 4.008 grams of Ca, therefore 4.008 grams of Ca will give us : $$ \frac{1}{20}\times4.008=0.2004\:g\:H_2 $$ Now, we know that the 0.2004 grams of hydrogen gas will be released. Let's calculate the number of moles in it : $$ =\frac{0.2004}{2}=0.1002\:moles $$ Also, we know that 1 mole of any gas occupies 22.7 L of volume at STP. So, 0.1002 moles of gas will occupy : $$ 0.1002\times22.7=2.27\:L $$ Therefore, 2.27 litres of hydrogen gas will be released.