Answer: 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]{\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles[/tex]
[tex]\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles[/tex]
[tex]8H_2+S_8\rightarrow 8H_2S[/tex]
According to stoichiometry :
1 mole of [tex]S_8[/tex] require = 8 moles of [tex]H_2[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] will require=[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2[/tex]
Thus [tex]S_8[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
As 1 mole of [tex]S_8[/tex] give = 8 moles of [tex]H_2S[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] give =[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2S[/tex]
Mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g[/tex]
Thus 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
