Calculate the approximate mas (in g ) of `H _(2)S` required for the following reaction when 15 L of oxygen at STP reacts completely.
`x H _(2)S(g) + yO_(2) (g) to aSO_(2)(g) + bH_(2)O (g)`
(molar mass of `H_(2)S =34.g mol ^(-1))`

Options

A. `12.11`
B. `15.16`
C. `34.12`
D. `68.12`

Answer

B

Explanation

`2H_(2)S + 3O _(2) to 2SO_(2) + 2H _(2)O`
From balanced equation
2 moles of `O _(2)` react with 2 moles of `H _(2)S`
`22.4 L of O _(2) =1` mol
`therefore 15 L of O _(2) = ((1)/(22.4) xx 15) mol`
3 moles of oxygen uses 2 moles of `H _(2)S.`
`therefore ((15)/(22.4))` moles of `O_(2)` uses `((2)/(3) xx (15)/(22.4))`moles of `H _(2)S`
Molar mass of `H_(2)S=34g//mol`
`therefore` Gram of `H_(2)S= (34 xx (2)/(3) xx (15)/(22.4))g = 15.17 g`