Question

For the following reaction, 3.07 grams of sulfur dioxide are mixed with excess water. The reaction yields 2.84 grams of sulfurous acid (H₂SO3).

sulfur dioxide (g) + water (L)→ sulfurous acid (H₂SO3) (g)

What is the theoretical yield of sulfurous acid (H₂SO3)? 3.93grams

What is the percent yield for this reaction?72.2%​

Answer 1

3.93 and 72.2%

Step-by-step explanation:

Theoretical Yield:

`3.07 \: g \: SO₂ * (1 \: mol \: SO₂)/(64.07g \: SO₂) * (1 \: mol \: H₂SO₃)/(1 \: mol \: SO₂) * (82.09 \: g \: H₂SO₃)/(1 \: mol \:H₂SO₃ ) = 3.933 \: g \: H₂SO₃`

Percent Yield:

`(2.84 \: g)/(3.993 \: g) * 100 = 72.2\%`