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finlep [7]
2 years ago
12

Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. if 4.40 mol of m

agnesium and 4.40 mol of hydrochloric acid are reacted, how many moles of magnesium chloride are produced?
Chemistry
1 answer:
Georgia [21]2 years ago
8 0
The balanced chemical equation for the above reaction is as follows ;
Mg + 2HCl —> MgCl2 + H2
The stoichiometry of Mg to HCl is 1:2
This means that 1 mol of Mg reacts with 2 mol of HCl
Equal amounts of both Mg and HCl have been added. One reagent is the limiting reactant and other reactant is in excess.
Limiting reactant is the reagent that is fully used up in the reaction and the amount of Product formed depends on the amount of limiting reactant present.
In this reaction if Mg is the limiting reactant, 4.40 moles of Mg should react with 4.40x2 -8.80 moles of HCl.
But only 4.40 moles of HCl present therefore HCl is the limiting reactant that reacts with 4.40/2 = 2.20 moles of Mg
Stoichiometry of HCl to MgCl2 is 2:1
Since HCl moles reacted -4.40 mol
Then MgCl2 moles formed are 4.40/2 = 2.20 mol of MgCl2
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2 years ago
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Which best describes the heat that you can feel on the engine of a running car?
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Calculate how many grams of the product form when 16.7 g of liquid bromine reacts with solid potassium. Assume that there is mor
salantis [7]

Answer: 49.7 grams of the product will be formed.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

{\text {Number of moles of bromine}}=\frac{16.7g}{80g/mol}=0.21moles  

Thus Br_2 acts as limiting reagent as it limits the formation of product as potassium is in excess.

2K(s)+Br_2(l)\rightarrrow 2KBr(s)

According to stoichiometry:

1 mole of Br_2 reacts to give 2 moles of KBr

Thus 0.21 moles of Br_2 will react to give=\frac{2}{1}\times 0.21=0.42 moles of KBr

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49.7 grams of the product will be formed.

4 0
2 years ago
Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.
olasank [31]
Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1 
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

d =  \dfrac{m}{V} \to V =  \dfrac{m}{d}

V =  \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)

<span>The solvent volume will be:
</span>
1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

<span>Then the mass of the solvent is:
</span>
m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

\omega =  \dfrac{m_1}{M_1*m_2}

\omega =  \dfrac{200}{132*0,821}

\omega =  \dfrac{200}{108,372}

\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark

_________________________________
_________________________________


<span>Another way to find the answer:
</span>
We have the following data: 

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4 
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol 


<span>Let's find the number of mols (n), let's see:

</span>n =  \dfrac{m_1}{M_1}

n = \dfrac{200}{132}

n \approx 1,5\:mol

Now, we apply all the data found to the formula of Molality, let us see:

\omega =  \dfrac{n}{m_2}

\omega =  \dfrac{1,5}{0,8}

&#10;\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark

I hope this helps. =)
7 0
2 years ago
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Answer : The disadvantage of using hydropower is Land can be flooded, displacing people and wildlife

Explanation :

Hydropower is a renewable source of energy. In hydropower, electricity is generated using the mechanical energy of moving water.

Since the kinetic energy is necessary to move the turbines, it is essential that the hydropower plant is present near a water source which has moving water.

It is the most efficient way of harnessing the energy from naturally available energy sources. It does not produce any waste. It does not pollute water or air.

Since hydropower needs kinetic energy of moving water, there is always a possibility of nearby land getting flooded. This would affect the people and wildlife of nearby areas.

Therefore the disadvantage of using hydropower is that land can be flooded, displacing people and wildlife


4 0
2 years ago
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