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Vanyuwa [196]
2 years ago
11

What volume of 3.00 M HCl in liters is needed to react completely (with nothing left over) with 0.750 L of 0.100 M Na2CO3?

Chemistry
1 answer:
Alla [95]2 years ago
5 0

Answer:

0.0500L of 3.00M HCl are needed to react completely

Explanation:

Hydrochloric acid HCl reacts with sodium carbonate, Na2CO3, as follows:

2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂

<em>Where 2 moles of HCl reacts per mole of Na₂CO₃</em>

<em />

To find the volume of 3.00M HCl needed to react completely we need first to calculate the moles of Na₂CO₃ added and, with this chemical equation we can solve for moles of HCl as follows:

<em>Moles Na₂CO₃:</em>

0.750L * (0.100moles Na₂CO₃ / L) = 0.0750moles

<em>Moles HCl:</em>

0.0750moles Na₂CO₃ * (2 moles HCl / 1 mole of Na₂CO₃) = 0.150 moles of HCl

<em>Volume of 3.00M HCl:</em>

0.150 moles of HCl * (1L / 3.00moles) =

<h3>0.0500L of 3.00M HCl are needed to react completely</h3>
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A 5.32 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.12 g fluorine, what
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Answer:

The mass of KF in the mixture is 2.77 gms.

Explanation:

Given;

Total weight of mixture (LiF+KF)=5.32gms

Let, mass of KF in the mixture = x gms

⟹ mass of LiF in mixture =(5.32-x)gms.

We know that :

Atomic weight of F=19gms.

Atomic weight of Li =7gms.

Atomic weight of K = 39 gms.

moles=mass/(molecular weight)

Thus, moles of KF=x/58

and moles of LiF = (5.97-x)/26LiF=(5.97−x)/26

Thus,

moles of F in KF=moles of KF=x/58 ---(1)

moles of F in LiF =moles of LiF= (5.32-x)/26---(2)

From (1) & (2),

Total moles of Fluorine

=(x/58)+((5.32-x)/26)

Hence,

total weight of Fluorine in sample = moles*Atomic weight

=((x/58)+((5.32-x)/26))*19gms.

=3.12 gms.---(given)

Now, solving the equation for x,

26x +(5.32*58)-58x

=3.12*58*26/19

22x=308.56-247.62

x=60.94/22

=2.77 gms. (Answer)

Thus, the mass of KF in the mixture is 2.77 gms.

8 0
3 years ago
Why couple should use a method of birth control
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Answer:

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Explanation:

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3 years ago
Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6
Sergeeva-Olga [200]

Explanation:

The given data is as follows.

 Weight of solute = 75.8 g,   Molecular weight of solute (toulene) = 92.13 g/mol,    volume = 200 ml

  • Therefore, molarity of toulene is calculated as follows.

      Molarity = \frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}

                    = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}

                    = 4.11 M

Hence, molarity of toulene is 4.11 M.

  • As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

   Molality = \frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}

             = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}

             = 8.6 m

Hence, molality of given toulene solution is 8.6 m.

  • Now, calculate the number of moles of toulene as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{75.8 g}{92.13 g/mol}

                             = 0.8227 mol

Now, no. of moles of benzene will be as follows.

     No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{95.6 g}{78.11 g/mol}

                             = 1.2239 mol

Hence, the mole fraction of toulene is as follows.

         Mole fraction = \frac{\text{moles of toulene}}{\text{total moles}}

                             = \frac{0.8227 mol}{(0.8227 + 1.2239) mol}

                             = 0.402

Hence, mole fraction of toulene is 0.402.

  • As density of given solution is 0.857 g/cm^{3} so, we will calculate the mass of solution as follows.

         Density = \frac{mass}{volume}

     0.857 g/cm^{3} = \frac{mass}{200 ml}      (As 1 cm^{3} = 1 g)

                      mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

      Mass % = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100

                   = \frac{75.8 g}{171.4 g} \times 100

                   = 44.22%

Therefore, mass percent of toulene is 44.22%.

8 0
3 years ago
What is 93,000,000 in scientific notation?
DanielleElmas [232]
I believe it is 9.3x10^7
5 0
3 years ago
For the reaction 2 s (s) + 3 o2 (g) → 2 so3 (g), how much so3 can be produced from 4 g o2 and excess s? 1. 0.25 mol so3 2. 0.13
igor_vitrenko [27]
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
8 0
3 years ago
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