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3 years ago

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In the following reaction, how many grams of oxygen will react with 10.47 grams of benzene (C6H6)? 2C6H6 + 1502 12CO2 + 6H2O The

molar mass of benzene is 78.1074 grams and that of O2 is 32 grams.

1 answer:

goldfiish [28.3K]3 years ago

8 0

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Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

crimeas [40]

Answer:

The answer to your question is:

a) 80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction CH4 + 2O2 ⇒ CO2 + 2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

16 g of CH4 ---------------- 2(32) g of O2

20 g -------------- x

x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .

CH4 + 2O2 ⇒ CO2 + 2H2O

15 g 22 g

16 g of CH4 ---------------- 64 g of O2

15 g of CH4 --------------- x

x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

CH4 + 2O2 ⇒ CO2 + 2H2O

64 g of O2 ------------------ 44 g of CO2

22 g of O2 ------------------ x

x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________

CH4 + 2O2 ⇒ CO2 + 2H2O

10.31 g 20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0

2 years ago

A baby weighs 7 pounds, 4 ounces at birth and 7 pounds, I ounce at discharge. What percent of weight did the baby lose

Aloiza [94]

Answer:

2.6%

Explanation:

As, 1 ounce (oz) = 0.0625 pounds (lb)

Therefore, weight of baby at discharge = 7 lb,1 oz = 7+0.0625 lb = 7.0625 lb

Since, 1 oz = 0.0625 lb

⇒ 4 oz = 4×0.0625 = 0.25 lb

Therefore, weight of baby at birth = 7 lb,4 oz = 7+0.25 lb = 7.25 lb

The <u>amount of weight lost</u> is equal to the difference of weight of the baby at birth and discharge.

Therefore, <u>weight lost</u> = 7.25 lb - 7.0625 lb = <u>0.1875 lb</u>

Now, the <u>percentage of weight lost</u> by the baby is given by the amount of weight lost divided by the weight of the baby at birth.

Therefore, <u>the percentage of weight los</u>t = weight lost ÷ weight at birth = 0.1875 lb ÷ 7.25 lb × 100 = <u>2.6% </u>

6 0

3 years ago

All atoms of an element have the same number of ________. electrons neutrons protons protons plus neutrons

Cloud [144]

Answer:

protons

Explanation:

An element, by definition, always has the same number of protons. Sodium, element 11, has 11 protons. Anything with 11 protons is a sodium atom, regardless of the number of neutrons, electrons, or politicians.

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Chuge%5Cmathfrak%5Cred%7BQuEsTioN%3A-%7D" id="TexFormula1" title="\huge\mathfrak\red{QuEsTio

sladkih [1.3K]

Explanation:

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6 0

2 years ago

How many moles of oxygen must be placed

Annette [7]

0.24 moles of oxygen must be placed in a 3.00 L container to exert a pressure of 2.00 atm at 25.0°C.

The variables given are Pressure, volume and temperature.

Explanation:

Given:

P = 2 atm

V = 3 litres

T = 25 degrees or 298.15 K by using the formula 25 + 273.17 = K

R = 0.082057 L atm/ mole K

n (number of moles) = ?

The equation used is of Ideal Gas law:

PV = nRT

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

Putting the values given for oxygen gas in the Ideal gas equation, we get

n = $\frac{2 x 3}{0.082157 x 298.15}$

= 0.24

Thus, from the calculation using Ideal Gas law it is found that 0.24 moles of oxygen must be placed in a container.

Ideal gas law equation is used as it tells the relation between temperature, pressure and volume of the gas.

3 0

2 years ago

Read 2 more answers