Respuesta :
1 mole of any gas under STP has volume 22.4 L.
Molar mass M(H2S) = 2M(H) + M(S)= 2*1.0 +32.1=34.1 g/mol
15g * 1mol/34.1g= 0.440 mol
0.440 mol*1L/22.4L= 9.85 L
The volume of 15.0 grams of H2S at STP is 9.85 L.
Molar mass M(H2S) = 2M(H) + M(S)= 2*1.0 +32.1=34.1 g/mol
15g * 1mol/34.1g= 0.440 mol
0.440 mol*1L/22.4L= 9.85 L
The volume of 15.0 grams of H2S at STP is 9.85 L.
Answer:
[tex]9.86 lH_ {2}S[/tex]
Explanation:
STP means standard conditions of temperature and pressure and under these conditions, one mole of any substance occupies a volume of 22.4 l.
We have [tex]15gH_ {2}[/tex] to calculate the number of mols you must calculate the molecular mass of the compound.
H 2 (1g / mol)
S 32 g / mol[tex](32+2) g / mol = 34g / mol H_ {2}S[/tex]
[tex]15g H_{2}S \cdot \frac{1 mol H_ {2}S}{34g H_ {2}S} = 0.44 mol H_ {2}S[/tex]
Now we calculate the volume occupied by [tex]0.44 mol H_ {2}S[/tex]
[tex]0.44 molH_ {2}S \cdot \frac{22.4 l H_ {2}S}{1mol H_ {2}S} = 9.86lH_ {2}S[/tex]
the volume of 15.0 g of [tex]H_{2}S[/tex] is 9.86l.
