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3 years ago

If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent?

1 answer:

5 0

K:

m=155g
M=39g/mol

n = 155g / 39g/mol ≈ 3,97mol

KNO₃:

m=122g
M=101g/mol

n = 122g/101g/mol = 1,21mol

2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent

KNO₃ is limiting reagent

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3 years ago

Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H

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Answer : The theoretical yield of water formed from the reaction is 0.54 grams.

Solution : Given,

Mass of HCl = 1.1 g

Mass of NaOH = 2.1 g

Molar mass of HCl = 36.5 g/mole

Molar mass of NaOH = 40 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of HCl and NaOH.

$\text{ Moles of }HCl=\frac{\text{ Mass of }HCl}{\text{ Molar mass of }HCl}=\frac{1.1g}{36.5g/mole}=0.030moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{2.1g}{40g/mole}=0.525moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NaOH

So, 0.030 mole of HCl react with 0.030 mole of NaOH

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