Question

Hydrogen sulfide gas and sulfur dioxide gas react to form solid sulfur and water. If 10.0g of gaseous hydrogen sulphide is mixed with 10.0g of gaseous sulfur dioxide, how many grams of solid sulfur can form?

Answer 1

\[ 2 H_2S + SO_2 → 3 S + 2 H_2O \]

\[(10.0 g H2S) / (34.0814 g H2S/mol) = 0.29342 mol H2S \\ (10.0 g SO2) / (64.0644 g SO2/mol) = 0.15609 mol SO2\]

0.29342 mole of H2S would react completely with 0.29342 x (1/2) = 0.14671 mole of SO2, but there is more SO2 present than that, so SO2 is in excess and H2S is the limiting reactant.

\[(0.29342\ mol \ H2S) x (3/2) * (32.0655 g \ S/mol) = 14.1\ g \ S \]

((0.15609 mol SO2 initially) - ( 0.14671 mol SO2 reacted)) x (64.0644 g SO2/mol) =
0.601 g SO2 left over

Answer 2

now i wonder how someone who didn't know compounds suddenly explains this huh @Rohangrr

Answer 3

http://au.answers.yahoo.com

Answer 4

anyway.... @Rohangrr 's equation is right...unfortunately the solution after isn't

Answer 5

I got frm there

Answer 6

@lgbasallote

Answer 7

@lgbasallote

Answer 8

yeah. it's wrong

Answer 9

anyway... first step is to find the limiting reagent

10 g of H2S would be

\[10 \; g \; H_2 S \times \frac{ 1 \; mol}{34.08 \; g} \times \frac {3 \; mol}{2 \; mol} = 0.440\; mol \]
10 g of SO2 would be
\[10 \; g \; SO_2 \times \frac{1 \; mol}{64.07 \; g} \times \frac{3 \; mol}{1\; mol} = 0.461\]

so therefore, H2S is the limiting reagent

so now... multiply the limiting reagent this by the molar mass of S

\[0.440 \; mol \ H_2 S \times \frac{32.07 \; g}{1\; mol} = 14.11 \; g\]

that should be the answer

Answer 10

do you get it @cmkc109 ?