Answer: a) 7.57 g
b) 4.97 L
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Sb_2S_3=\frac{25.0g}{339.7g/mol}=0.074moles[/tex]
[tex]Sb_2S_3+6HCl\rightarrow 3H_2S+2SbCl_3[/tex]
As [tex]HCl[/tex] is the excess reagent, [tex]Sb_2S_3[/tex] is the limiting reagent and it limits the formation of product.
According to stoichiometry :
1 mole of [tex]Sb_2S_3[/tex] form = 3 moles of [tex]H_2S[/tex]
Thus 0.074 moles of [tex]Sb_2S_3[/tex] will form = [tex]\frac{3}{1}\times 0.074=0.222moles[/tex] of [tex]H_2S[/tex]
Mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.222moles\times 34.1g/mol=7.57g[/tex]
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = ?
n = number of moles = 0.222
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]273K[/tex]
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{0.222\times 0.0821L atm/K mol\times 273K}{1}=4.97L[/tex]
Thus the volume occupied by [tex]H_2S[/tex] is 4.97 L
