Question

Suppose hydrochloric acid reacts with potassium sulfite yielding water, sulfur dioxide, and potassium chloride. Suppose 4 moles of hydrochloric acid react with excess potassium sulfite. How many grams of sulfur dioxide are produced?

Answer 1

Answer:

128 grams of sulfur dioxide are produced.

Explanation:

$2HCl+K_2SO_3\rightarrow SO_2+2KCl+H_2O$

Moles of HCl = 4 moles

According to reaction, 2 moles of HCl gives 1 mole of sulfur dioxide gas.

Then 4 moles of HCl will give:

$\frac{1}{2}\times 4 moles=2 moles$ of sulfur dioxide gas.

Mass of sulfur dioxide gas = 2 mol × 64 g/mol = 128 g

128 grams of sulfur dioxide are produced.

Answer 2

The amount, in grams, of sulfur dioxide that would be produced will be 128 grams

Stoichiometric calculation

From the balanced equation of the reaction:

$2HCl + K_2SO_3 --- > H_2O + SO_2 + 2KCl$

Mole ratio of HCl and SO2 = 2:1

Thus, 4 moles of HCl will produce 2 moles of SO2

Mass of 2 moles SO2 = 2 x 64

                                  = 128 grams

More on stoichiometric calculations can be found here: https://brainly.com/question/8062886