If you need 2.5 moles of Rb2Cr7, how many grams do you need?

The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the

Svetach [21]

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and $NH_3$ .

$\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles$

$\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl$

From the balanced reaction we conclude that

As, 1 mole of $NaCl$ react with 1 mole of $NH_3$

So, 0.243 mole of $NaCl$ react with 0.243 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and $NaCl$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaHCO_3$

From the reaction, we conclude that

As, 1 mole of $NaCl$ react to give 1 mole of $NaHCO_3$

So, 0.243 moles of $NaCl$ react to give 0.243 moles of $NaHCO_3$

Now we have to calculate the mass of $NaHCO_3$

$\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3$

Molar mass of sodium bicarbonate = 84 g/mol

$\text{ Mass of }NaHCO_3=(0.243moles)\times (84g/mole)=20.4g$

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

4 0

3 years ago

During which lunar phase is the moon’s near side entirely lit?

Bas_tet [7]

<span> full moon is the best answer, but perhaps 90% lit, not entirely. The features we see on a full moon are shadows = not lit areas.</span>

5 0

3 years ago

Calculate the vapor pressure (in torr) of a 35°C solution of ethanol and propanol where the mole fraction of propanol is .75. As

ASHA 777 [7]

The vapor pressure (in torr) of a 35°C solution : 52 torr

<h3>Further explanation</h3>

Given

x propanol = 0.75

P°ethanol=100 torr

P°propanol=36 torr

Required

the vapor pressure

Solution

Raoult's Law

P solution = xA.P°A +xB.P°B

x ethanol = 1- 0.75 = 0.25

Input the value :

P solution = 0.25 x 100 + 0.75 x 36

P solution = 25 + 27

P solution = 52 torr

7 0

2 years ago

These are incorrect please put the right answer on the blanks\<br><br>for brainiest:)

harina [27]

Answer:

1. condensation

2. Temperature

3. evaporization

4. kinetic energy

5. sub limation

6. thermal energy

7. deposition

8. vaporization

3 0

2 years ago

Need little help pls

Julli [10]

Answer:

0.354 molal

Explanation:

The molarity (M) or the concentration of a solution is defined as the number of moles of a compound in the solution per liter of solution (mol/L), whereas molality, is defined as the number of moles of a compound in the solution per kg of the compound (mol/kg).

Given that the density of the solution is 1.202 g/mL, which is equivalent to 1.202 kg/L. Since the prefix mili- denotes a factor of one thousandth ( $10^{-3}$ ) and kilo- denotes a factor of one thousand ( ${10}^3$ ),

$1.202 \ \frac{g}{mL} \ = \ 1.202 \ \frac{g}{(10^{-3}) L} \ = \ 1.202 \ \frac{{10}^3g}{L} \ = \ 1.202 \ \frac{kg}{L}$ .

To calculate the corresponding molality of the solution, the formula

$Molality \ (mol/kg) \ = \ \frac{Molarity \ (mol/L)}{Density \ (kg/L)}$ is used.

Therefore,

$Molality \ = \ \frac{0.426 \ mol/L}{1.202 \ kg/L} \ = \ 0.354 \ mol/kg \ \ (3 \ s.f.)$

4 0

2 years ago