Question

Two samples of potassium iodide are decomposed into their constituent elements. The first sample produced 13.0 g of potassium and 42.3 g of iodine. If the second sample produced 24.4 kg of potassium, how many kg of iodine were produced?

Answer 1

Answer:

The mass of iodine produced will be 79.46 kg .

Explanation:

Let us calculate the mole ratio of potassium and iodine formed when first sample was decomposed.

Atomic mass of potassium =39 g/mol

Atomic mass of iodine = 127 g/mol

Moles of K produced = $\frac{mass}{atomicmass}=\frac{13}{39}= 0.33$

Moles of I produced =  $\frac{mass}{atomicmass}=\frac{42.3}{127}= 0.33$

The mole ration of K :I = 1:1

The moles of K produced by second sample = $\frac{mass}{molarmass}=\frac{24.4X10^{3}}{39}= 0.626X10^{3} mol$

The same moles of I must have been produced.

The mass of Iodine produced = moles X atomic mass = 0.626  X 127=79.46 kg

Answer 2

79.46kg of Iodine will be produced in the reaction.

What is an Element?

These are substances which can't be broken down into smaller matter and is usually very pure.

Parameters

Atomic mass of potassium =39 g/mol

Atomic mass of iodine = 127 g/mol

Moles of K produced = ?

Moles of I produced =  ?

The moles of K produced = mass/molar mass

= 13/39

= 0.33

Moles of I produced=  mass/molar mass

=42.3/127

= 0.33

Mole ratio of K:I = 1:1

The moles of K produced by second sample = mass/molar mass =

= 24.4 × 10³ / 39 = 0.626mol

The mass of Iodine produced = moles X atomic mass

= 0.626  X 127= 79.46 kg

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