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3 years ago

5. What is the molar mass of ethanol (C2H5OH)?

Chemistry

1 answer:

AVprozaik [17]3 years ago

5 0

The molar mass is given by the sum of the atomic masses of the component elements of the substance.

In our case, the substance is ethanol or C₂H₅OH.

$A_C=12.011$

$A_H=1.0079$

$A_O=15.9994$

$\mathcal M_{C_2H_5OH} = 2 \times A_C + 6 \times A_H + A_O = 2 \times 12.011 + 6 \times 1.00079+ 15.999 4= 24.022 + 6.0474 + 15.9994= 46.069≈ \boxed{46.07 \dfrac{g}{mol}}$

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Help me out? thank you.

Vladimir [108]

Wow very difficult I am so confused

3 0

3 years ago

Read 2 more answers

for the following reaction how many grams of h20 will be produced if you react 3.7 grams of B2H6? B2h6 +3O2-2hbo2+ 2 H20

Nana76 [90]

Answer:

4.8 g H₂O

Explanation:

To find the mass of water, you need to (1) convert grams B₂H₆ to moles B₂H₆ (via molar mass from periodic table), then (2) convert moles B₂H₆ to moles H₂O (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles H₂O to grams H₂O (via molar mass from periodic table).

It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs because the given value (3.7 grams) has 2 sig figs.

Molar Mass (B₂H₆): 2(10.811 g/mol) + 6(1.008 g/mol)

Molar Mass (B₂H₆): 27.67 g/mol

1 B₂H₆ + 3 O₂ ---> 2 HBO₂ + 2 H₂O
^ ^

Molar Mass (H₂O): 15.998 g/mol + 2(1.008 g/mol)

Molar Mass (H₂O): 18.014 g/mol

3.7 g B₂H₆ 1 mole 2 moles H₂O 18.014 g
---------------- x --------------- x ----------------------- x ----------------- = 4.8 g H₂O
27.67 g 1 mole B₂H₆ 1 mole

6 0

2 years ago

Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/

Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of $C_2H_4$ and $H_2$ = 25.0 atm

Temperature of $C_2H_4$ and $H_2$ = $250^oC=273+250=523K$

Volume of $C_2H_4$ = 1050 L per min

Volume of $H_2$ = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of $C_2H_6$ = 30 g/mole

First we have to calculate the moles of $C_2H_4$ and $H_2$ by using ideal gas equation.

For $C_2H_4$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=611.34moles$

For $H_2$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=902.46moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_2H_4+H_2\rightarrow C_2H_6$

From the balanced reaction we conclude that

As, 1 mole of $C_2H_4$ react with 1 mole of $H_2$

So, 611.34 mole of $C_2H_4$ react with 611.34 mole of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $C_2H_6$ .

As, 1 mole of $C_2H_4$ react to give 1 mole of $C_2H_6$

As, 611.34 mole of $C_2H_4$ react to give 611.34 mole of $C_2H_6$

Now we have to calculate the mass of $C_2H_6$ .

$\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6$

$\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g$

The theoretical yield of $C_2H_6$ = 18340.2 g

The actual yield of $C_2H_6$ = 14.0 kg = 14000 g (1 kg = 1000 g)

Now we have to calculate the percent yield of $C_2H_6$

$\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%$

Therefore, the percent yield of the reaction is, 76.34 %

5 0

3 years ago

You have a different rock with a volume of 30cm and a mass of 60g, What is its density?

Galina-37 [17]

Density = mass/volume

Therefore,
Density = 60g/30cm

8 0

3 years ago

Write molecular, complete ionic, and net ionic equations for the reactions that occur, if any, when solutions of the following s

Gnom [1K]

<span>The molecular formula that describes the problem is
2CH3COOH (aq) + Ca(OH)2 (s) ---> Ca(CH3COO)2 (aq) + 2H2O (l)

The net equation is written as follows:
2CH3COOH- (aq) + 2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 CH3COO- (aq) + 2H2O (l)
canceling out spectator ions
2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 H2O (l)</span>

4 0

3 years ago