CHEMISTRY COLLEGE

how many grams of F2 are needed to react with 3.50 grams of Cl2 Equation needed for question- Cl2+3F2-->2ClF3 please explain how to get the answer.

Answers

Answer 1

Answer:

Answer:

5.62 g of F2

Explanation:

We have to start with the chemical reaction:

$Cl_2~+~3F_2~-->~2ClF_3$

We have a balanced reaction, so we can continue with the mol calculation. For this, we need to know the molar mass of $Cl_2$ (70.906 g/mol), so:

$3.5~g~Cl_2(1~mol~Cl_2)/(70.906~g~Cl_2)=0.049~mol~Cl_2$

Now, with the molar ratio between $Cl_2$ and $F_2$ we can convert from moles of $Cl_2$ and $F_2$ (1:3), so:

$0.049~mol~Cl_2(3~mol~F_2)/(1~mol~Cl_2)=0.148~mol~F_2$

Finally, with the molar mass of $F_2$ we can calculate the gram of $F_2$ (37.99 g/mol), so:

$0.148~mol~F_2(37.99~g~F_2)/(1~mol~F_2)=5.62~g~F_2$

I hope it helps!

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A solution containing 292 g of Mg(NO3)2 per liter has a density of 1.108 g/mL. The molality of the solution is:A) 2.00 m

B) 1.77 m

C) 6.39 m

D) 2.41 m

E) none of these

Answers

Answer: D) 2.41 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molality=(n)/(W_s)$

where,

n = moles of solute

$W_s$ = weight of solvent in kg

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}=(292g)/(148g/mol)=1.97moles$

volume of solution = 1L = 1000 ml (1L=1000ml)

Mass of solution= ${\text {Density of solution}}* {\text {Volume of solution}}=1.108g/ml* 1000ml=1108g$

mass of solute = 292 g

mass of solvent = mass of solution - mass of solute = (1108- 292) g = 816g = 0.816 kg

Now put all the given values in the formula of molality, we get

$Molality=(1.97moles)/(0.816kg)=2.41mole/kg$

Therefore, the molality of solution will be 2.41 mole/kg

Final answer:

In this problem, we calculate molality by using the given mass of the solute, the mass of the solvent, and the molar mass of the solute. After performing the necessary calculations, we find that the molality is 2.41 m.

Explanation:

The subject of this student's question is molality, which is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. To find the molality (m), we need to know the mass of the solute and the mass of the solvent in the solution.

Given, that the solution contains 292g of Mg(NO3)2 per liter (which is the mass of the solute). The density of the solution is 1.108g/mL. We know that 1L = 1000mL, so the mass of the solution is density x volume = 1.108g/mL x 1000mL = 1108g.

We need to find the mass of the solvent (water). The mass of the solution is the mass of the solute + the mass of the solvent. So, the mass of the solvent is 1108g(mass of the solution) - 292g(mass of solute) = 816g or 0.816gkg.

The molar mass of Mg(NO3)2 is 148.31452 g/mol. So, the number of moles of Mg(NO3)2 in the solution is moles = mass / molar mass = 292g / 148.31452 g/mol = 1.97 moles.

Now we can calculate molality (m) = moles of solute/mass of solvent in kg = 1.97 moles / 0.816 kg = 2.41 m. Therefore, the answer is D) 2.41 m.