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zvonat [6]
3 years ago
6

Can the excess reactant (in stoichiometry) be negative?

Chemistry
1 answer:
strojnjashka [21]3 years ago
6 0
No. Think about it this way: it’s a physical, leftover substance, you can never have a negative amount of anything.

If you’re speaking in terms of charge, however, then yes, it’s very likely.
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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
Pie

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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An engineer uses aluminum to build an airplane rather than composite
lesantik [10]

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A

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Can unit rate represent how much for one unit
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What is the molality of a solution made by combining 5.0 g of CaCl2 with 500.0 g of water?
ankoles [38]
Molality is defined as the number of moles of solute dissolved in 1 kg of solvent.
To calculate molality, we need to calculate the number of moles of CaCl₂.
Mass of CaCl₂ - 5.0 g
Molar mass of CaCl₂ - 111 g/mol
The number of moles of CaCl₂ - 5.0 g / 111 g/mol = 0.045 mol 
we need to then calculate the number of moles in 1 kg solvent.
number of CaCl₂ moles in 500 g water - 0.045 mol
Therefore number of moles in 1 kg water - 0.045 mol / 500g x 1000 g = 0.090 mol
Molality of CaCl₂ - 0.090 mol/kg

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3 years ago
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