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Grace [21]
3 years ago
10

A reaction produces 74.10 g Ca(OH)2 after 56.08 g CaO is added to 36.04 g H2O. How should the difference in the masses of reacta

nts and products be explained?
Chemistry
1 answer:
melomori [17]3 years ago
3 0
Cao +  H2O  ---->Ca(OH)2
Calculate   the  number  of  each   reactant  and  the  moles  of  the  product
that  is
moles = mass/molar mass
The  moles  of  CaO=  56.08g/  56.08g/mol(molar  mass  of  Cao)=  1mole
the  moles  of  water=  36.04 g/18  g/mol=  2.002moles
The   moles  of Ca (OH)2=74.10g/74.093g/mol= 1mole

 The  mass  of differences  of  reactant  and  product  can   be  therefore 
 explained  as 
 1  mole   of  Cao  reacted  completely   with   1  mole   H2O  to  produce  1 mole  of  Ca(OH)2. The  mass  of  water   was  in  excess  while  that  of  CaO  was  limited

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<u>Answer:</u> The mass of decane produced is 1.743\times 10^2g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:  

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol

The chemical equation for the hydrogenation of decene follows:

C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = \frac{1}{1}\times 1.225=1.225mol of decane

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Moles of decane = 1.225 mol

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Putting values in equation 1, we get:

1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g

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