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Paladinen [302]
3 years ago
12

How many grams are in 2.40 moles of sodium hydroxide (NaOH)?

Chemistry
1 answer:
Bogdan [553]3 years ago
5 0

Answer:

96g

Explanation:

Given parameters:

Number of moles of NaOH = 2.4moles

Unknown:

Mass of NaOH = ?

Solution:

The mass of a substance given the number of moles can be found using the expression below;

     Mass of NaOH  = Number of moles x molar mass;

  Molar mass of NaOH  = 23 + 16 + 1 = 40g/mol

Now input the parameters and solve;

    Mass of NaOH  = 2.4 x 40  = 96g

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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

\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g

Therefore, the theoretical yield of HCl is, 56.1735 grams

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