Calculate the number of moles of excess reactant that will be left-over when 50.0 g of ki react with 50.0 g of br2: 2ki + br2 2kbr + i2

Answer : The moles of excess reactant, is, 0.1625 mole.

Explanation : Given,

Mass of = 50 g

Mass of = 50 g

Molar mass of = 166 g/mole

Molar mass of = 160 g/mole

First we have to calculate the moles of and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 moles of react with 1 mole of

So, 0.301 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

The moles of excess reactant, = Given moles - Required moles

The moles of excess reactant, = 0.313 - 0.1505 = 0.1625 mole

Therefore, the moles of excess reactant, is, 0.1625 mole.

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chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.