Answer:
[tex]m_{HCl}=36.1gHCl[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:
[tex]n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2[/tex]
Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:
[tex]m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl[/tex]
Regards!
