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2 years ago

4

Isopropanol (C3H8O) is a key ingredient in some hand sanitizers. Suppose that 127 grams of isopropanol is dissolved in water. Th

e volume of the solution is 1,250 milliliters. What is the molarity of the solution? Refer to the periodic table to help you answer. Express your answer to three significant figures. The molarity of the solution is M.

2 answers:

sweet [91]2 years ago

8 0

Answer:- 1.69M

Solution:- Molarity is moles of solute per liter of solution. We have 127 grams of the solute, isopropanol and the volume of the solution is 1250 mL. We need to convert the grams to moles and the mL to L.

Molar mass of $C_3H_8O$ = 3(12.011) + 8(1.008) + 1(15.999) = 60.096 gram per mol

Let's convert the grams to moles as:

$127gC_3H_8O(\frac{1mol}{60.096g})$

= 2.11 mol

The volume is 1250 mL that is 1.25 L.

$molarity=\frac{2.11mol}{1.25L}$

= 1.69M

So, the molarity of the solution is 1.69M.

posledela2 years ago

7 0

Answer: 1.69 M

Explanation:

1) Data:

a) mass of solute: 127 g

b) volume of solute: 1,250 ml = 1.25 liters

c) compound: C₃H₈O

d) M = ?

2) Formulae:

a) M = number of moles of solute / volume of solution in liters

b) number of moles = mass in grams / molar mass

3) Solution

a) molar mass of C₃H₈O:

element number of atoms atomic mass mass of the element

a.m.u. g / mol

C 3 12.011 3 × 12.011 = 36.033

H 8 1.008 8 × 1.008 = 8.064

O 1 15.999 15.999

---------------------------------

Sum = molar mass = 60.096 g/mol

b) number of moles = mass in grams / molar mass = 127 g / 60.096 g/mol = 2.113 mol

c) M = number of moles of solute / volume of solution in liters = 2.113 mol / 1.250 liter = 1.69 M (showing 3 significant figures).

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Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix

Genrish500 [490]

Answer :

(a) Limiting reactant = $LiH$

(b) The excess reactant = $BCl_3$

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

(e) The mass of $LiCl$ produced is, 1066.42 lb

Explanation : Given,

Mass of $LiH$ = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of $BCl_3$ = 1000 lb = 453592 g

Molar mass of $LiH$ = 7.95 g/mole

Molar mass of $BCl_3$ = 117.17 g/mole

Molar mass of $B_2H_6$ = 27.66 g/mole

Molar mass of $LiCl$ = 42.39 g/mole

First we have to calculate the moles of $LiH$ and $BCl_3$ .

$\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles$

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$6LiH+2BCl_3\rightarrow B_2H_6+6LiCl$

From the balanced reaction we conclude that

As, 6 moles of $LiH$ react with 1 mole of $BCl_3$

So, 11411.13 moles of $LiH$ react with $\frac{11411.13}{6}=1901.855$ moles of $BCl_3$

From this we conclude that, $BCl_3$ is an excess reagent because the given moles are greater than the required moles and $LiH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant $(BCl_3)$ .

$\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100$

$\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%$

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of $B_2H_6$ .

As, 6 moles of $LiH$ react to give 1 mole of $B_2H_6$

So, 11411.13 moles of $LiH$ react to give $\frac{11411.13}{6}=1901.855$ moles of $B_2H_6$

Now we have to calculate the mass of $B_2H_6$ .

$\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6$

$\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g$

Now we have to calculate the percent yield of $B_2H_6$ .

$\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%$

The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

Now we have to calculate the moles of $LiCl$ .

As, 6 moles of $LiH$ react to give 6 mole of $LiCl$

So, 11411.13 moles of $LiH$ react to give 11411.13 moles of $LiCl$

Now we have to calculate the mass of $LiCl$ .

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

$\text{Mass of }LiCl=(11411.13mole)\times (42.39g/mole)=483717.8007g=1066.42lb$

The mass of $LiCl$ produced is, 1066.42 lb

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In selected option, HCl is reacting as Acid as it donates H⁺ to water (lowery bronsted base).

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