Question

What is the volume of 15.0 grams of H2S at STP

Answer 1

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.

Answer 2

Answer:

$9.86 lH_ {2}S$

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 $15gH_ {2}$ to calculate the number of mols you must calculate the molecular mass of the compound.

H 2 (1g / mol)

S 32 g / mol$(32+2) g / mol = 34g / mol H_ {2}S$

$15g H_{2}S \cdot \frac{1 mol H_ {2}S}{34g H_ {2}S} = 0.44 mol H_ {2}S$

Now we calculate the volume occupied by $0.44 mol H_ {2}S$

$0.44 molH_ {2}S \cdot \frac{22.4 l H_ {2}S}{1mol H_ {2}S} = 9.86lH_ {2}S$

the volume of 15.0 g of $H_{2}S$ is 9.86l.