How many moles of sodium chloride (NaCl) solute are in 155 grams of an 85.5 percent by mass solution?

Chemistry

2 answers:

Ira Lisetskai [31]3 years ago

8 0

3.0 L*3.5mol/ 1 L dillute solutio = 10.5 mol HCl

10.5mol*1 L stock solution/18.5 mol = .568L stock solution

3.0L - .568 L = 2.93 liters of water added

Measure .727 L of the 18.5 molar HCl stock solution, then add 2.93 of water to make 3.0 K of 3.5 M solution.

pogonyaev3 years ago

4 0

How many moles of sodium chloride (NaCl) solute are in 155 grams of an 85.5 percent by mass solution?

1) Calculate the mass of solute

% = [mass of solute / mass of solution] * 100 =>

mass of solute = [% / 100] * mass of solution = [85.5/100] * 155 g = 132.525 g

2) Pass to moles using the molar mass, MM, of NaCl

MM = 23 g/mol + 35.5 g/mol = 58.5 g/mol

n = mass (g) / MM = 132.525 / 58.5 = 2.27 mol

Answer: the second option of the list

4. How many grams of zinc metal will react completely with 8.2 liters of 3.5 M HCl? Show all of the work needed to solve this problem.

Zn (s) + 2 HCl (aq) ---> ZnCl2 (aq) + H2 (g)

From the molarity, calculate the number of mols of HCl

M = n/V => n = M*V = 3.5 mol/liter * 8.2 liter = 28.7 mol HCl

From the reaction: [1 mol Zn / 2 mol HCl] * 28.7 mol HCl = 14.35 mol Zn

Pass to mass, using the atomic mass of Zn

mass = n*atomic mass = 14.35 mol * 65.4 g/mol = 938.5 g Zn

Answer: 938.5 g.

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When 0.100 mol of carbon is burned in a closed vessel with8.00

antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of $O_2$ = 8.00 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of $O_2$ .

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C+O_2\rightarrow 2CO$

From the balanced reaction we conclude that

As, 2 mole of $C$ react with 1 mole of $O_2$

So, 0.1 moles of $C$ react with $\frac{0.1}{2}=0.05$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO$