For the following reaction, 101 grams of magnesium nitride are allowed to react with 144 grams of water. mg3n2 (5) + 6 h20 (1) — 3 mg(oh)2 (aq) + 2 nh2 (aq) what is the formula for the limiting reagent? what is the maximum amount of magnesium hydroxide that can be formed? grams what amount of the excess reagent remains after the reaction is complete? grams submit answer retry entire group 8 more group attempts remaining

Answer : The formula of limiting reagent is, .

The mass of is, 174 grams.

The mass of excess reactant is, 36 grams.

Solution : Given,

Mass of = 101 g

Mass of = 144 g

Molar mass of = 101 g/mole

Molar mass of = 18 g/mole

Molar mass of = 58 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, 1 mole of react with 6 mole of

So, given 1 moles of react with 6 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.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 1 mole of react to give 3 mole of

So, given 1 mole of react to give 3 moles of

Now we have to calculate the mass of

The mass of is, 174 grams.

Now we have to calculate the moles of excess reactant .

Moles of excess reactant = 8 - 6 = 2 moles

Now we have to calculate the mass of excess reactant.

The mass of excess reactant is, 36 grams.

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