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3 years ago

72.0 grams of water how many miles of sodium with react with it?

Chemistry

1 answer:

Flura [38]3 years ago

7 0

Answer:

$\large \boxed{\text{8.00 mol}}$

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

Mᵣ: 18.02

2Na + H₂O ⟶ 2NaOH + H₂

m/g: 72.0

2. Moles of H₂O

$\text{Moles of H$_{2}$O} = \text{72.0 g H$_{2}$O} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g H$_{2}$O}} = \text{3.996 mol H$_{2}$O}$

3. Moles of Na

The molar ratio is 2 mol Na/1 mol H₂O.

$\text{Moles of Na} = \text{3.996 mol H$_{2}$O} \times \dfrac{\text{2 mol Na}}{\text{1 mol H$_{2}$O}} = \textbf{8.00 mol Na}\\\\\text{The water will react with $\large \boxed{\textbf{ 8.00 mol}}$ of Na}$

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2 years ago

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To study a key fuel-cell reaction, a chemical engineer has 20.0-L tanks of H₂ and of O₂ and wants to use up both tanks to form 2

Lapatulllka [165]

the pressure needed on H2 tank is 34.9477319atm and the pressure needed in O2 tanks is 16.7690007atm .

Given , to study a key fuel-cell reaction , a chemical engineer has 20.0L tanks of H2 and O2 and wants to use up both tanks to form 28.0mol of water at 23.8°C .

the reaction of the fuel cell is given by ,

H2(g) +1/2 O2 ( g) →H2O

Moles of H2 required = 28 mol

moles of O2 required = 14mol

Now according to van der waals equation ,

p= (nRT/ V-nb) - an^2 / V^2

for H2 , a = 0.2453L^2bar /mol^2 , b= 0.02651L/mol

P for H2

= (28×0.0821×296.8/20-28×0.02651 ) -(0.2456 ×28×28/20×20)

P for H2 = 35.4291079-0.481376 = 34.9477319 atm

for O2 , a = 1.38L^2bar/mol^2 , b = 0.03186L/mol

P for O2 = (14×0.0821×296.8 /20-14×0.03186) - (1.382×14×14/20×20)

P for O2 = 17.4461807 - 0.67718 =16.7690007 atm

Hence , the pressure needed on H2 tank is 34.9477319atm and the pressure needed in O2 tanks is 16.7690007atm .

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11 months ago

You've made the hypothesis that the higher the density of a rock, the higher the temperature at which will melt, because a dense

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3 years ago

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a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (<em>Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol</em>), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: <em>Moles solution = moles solute + moles solvent.</em> First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

$Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03$

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3 years ago