Answer to Question #71561 in General Chemistry for Randrika

Question

Determine the number of gram H2 froemed when 250.0mL of 0.743 M HCL solution reacts with 3.41 x 10^23 atoms of FE according to the following reaaction. 2HCL(aq) + Fe(s) yeild H2(g) + FECL2(aq)

Answer 1

n (HCl) = 250ml*0.743M=185.75 mmole
n (Fe)=N/Na=(3.41*10^23)/(6.02*10^23)=0.56645 mole=566.45 mmole
n(HCl)<n(Fe)=>HCl is limiting reagent
n(H2)=n(HCl)/2=185.75 mmole/2=92.875 mmole
m (H2)=n(H2)*m(H2)=0.92875 mole*2 g/mole=0.186 g