Question

For the following reaction, if i start this reaction with 153.0 grams of hydrobromic acid and an excess of potassium bi sulfate, a) how many grams of sulfur dioxide gas will I produce? b) if 98.75 g are actually achieved, what is the % yield of the reaction? HBr(aq) + KHSO3(aq) --> KBr(aq)+SO2(g)+H2O(I)  

Answer 1

a)

Balanced reaction

HBr (aq) + KHSO₃ (aq) $\to$ KBr (aq) + SO₂ (g) + H₂O (g)

mole ratio of HBr and SO₂ = 1:1

now moles of HBr used

= mass of HBr / molar mass of HBr

= 153.0 (g)  $\times$_{$\small \frac{1 (mole)}{80.91(g)}$}

= 1.89

therefore moles of SO₂ can be formed = 1.89

mass of SO₂ = moles* molar mass

= 1.89 (mol) $\times$ 64 ($\small \frac{g}{mol}$ )

= 121.02 g

b)

Percent yield

= $\small \frac{actual\, yield\times 100}{theoretical \, yield}$

= $\small \frac{98.75\times 100}{121.02}$

= 81.59 %