Question

Use the equation to answer the question.H2SO4+HI→H2S+I2+H2O Which option shows a correctly balanced chemical equation?(1 point)ResponsesH2SO4+8HI→H2S+4I2+4H2Oh sub 2 S O sub 4 plus 8H I rightarrow h sub 2 s plus 4i sub 2 plus 4h sub 2 oH2SO4+HI→H2S+I2+H2Oh sub 2 S O sub 4 plus H I rightarrow h sub 2 s plus i sub 2 plus h sub 2 oH2SO4+8HI→H2S+8I2+4H2Oh sub 2 S O sub 4 plus 8H I rightarrow h sub 2 s plus 8i sub 2 plus 4h sub 2 o2H2SO4+2HI→2H2S+I2+4H2O

Answer 1

To balance the equation \( \text{H}_2\text{SO}_4 + \text{HI} \rightarrow \text{H}_2\text{S} + \text{I}_2 + \text{H}_2\text{O} \), we need to ensure that the number of each type of atom is the same on both sides of the equation.

1. Counting the atoms:

   • Left side:
     • H: 2 (from H2SO4) + 1 (from HI) = 3
     • S: 1 (from H2SO4)
     • O: 4 (from H2SO4)
     • I: 1 (from HI)
   • Right side:
     • H: 2 (from H2S) + 2 (from H2O) = 4
     • S: 1 (from H2S)
     • O: 1 (from H2O)
     • I: 2 (from I2)

2. Balancing the equation:

   • Start with S, which is already balanced.
   • Balance I and H by adjusting the coefficients:
   • Following the original reaction, we can start with 8 HI to balance the iodine.
     • This leads to new counts (portray this as \( \text{H}_2\text{SO}_4 + 8 \text{HI} \)):
     • Left side: H (2 + 8) = 10, S (1), O (4), I (8)
     • Right side coming from \( \text{H}_2\text{S} + 4 \text{I}_2 + 4 \text{H}_2\text{O} \)
     • Right side: H (2 + 4) = 6, S (1), O (4), I (8)
   • So the balanced reaction becomes: \[ \text{H}_2\text{SO}_4 + 8 \text{HI} \rightarrow \text{H}_2\text{S} + 4 \text{I}_2 + 4 \text{H}_2\text{O} \]

Thus, the correctly balanced chemical equation from the options you provided is:

H2SO4 + 8HI → H2S + 4I2 + 4H2O.