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3 years ago

How many molecules are in 145.5 grams of Be(OH)2

Chemistry

1 answer:

Zanzabum3 years ago

3 0

Answer:

2.04 x 10²⁴ molecules

Explanation:

Given parameters:

Mass of Be(OH)₂ = 145.5g

To calculate the number of molecules in this mass of Be(OH)₂ we follow the following steps:

>> Calculate the number of moles first using the formula below:

Number of moles = mass/molarmass

Since we have been given the mass, let us derive the molar mass of Be(OH)₂

Atomic mass of Be = 9g

O = 16g

H = 1g

Molar Mass = 9 + 2(16 + 1)

= 9 + 34

= 43g/mol

Number of moles = 145.5/43 = 3.38mol

>>> We know that a mole is the amount of substance that contains Avogadro’s number of particles. The particles can be atoms, molecules, particles etc. Therefore we use the expression below to determine the number of molecules in 3.38mol of Be(OH)₂:

Number of

molecules= number of moles x 6.02 x 10²³

Number of molecules= 3.38 x 6.02 x 10²³

= 20.37 x 10²³ molecules

= 2.04 x 10²⁴ molecules

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Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

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Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$