All categories

3 years ago

If you have 600 g of c6h12o6 then how many grams of oxygen are there

1 answer:

7 0

so your ratio for the equation is 1 glucose + 6 oxygen = 6 carbon dioxide + 6 water

those ratios will all be in moles, and will all be based on the number of moles of your limiting reagent. We assume that you have unlimited oxygen, so glucose must be limiting. The molar mass for glucose is 180 grams (you can get more accurate if you need, this is just the number I am using).

100 g / 180 g/mol = 0.556 moles of glucose.
since we will need six times the moles of glucose in oxygen...
6 x .556 = 3.33 moles of O2
multiply by the molar mass of O2 for the mass of oxygen needed...
3.33 x 32 = 106.67 grams of O2
since you are forming six times the moles of glucose in CO2 and H2O, you are forming 3.33 moles of each of these compounds as well.

You might be interested in

A certain mass of carbon reacts with 9.53 g of oxygen to form carbon monoxide. ________ grams of oxygen would react with that sa

yawa3891 [41]

Answer: 9.53 *2= 19.06

Explanation:

The law of multiple proportions states that if two elements combines to form more than one compound the ratio of masses of the second element which combines to the fixed mass of the first element will always be the ratios of the small whole numbers.

in case of carbon monoxide, mass of carbon will be the same of mass of oxygen.

But in case of carbon dioxide, if carbon is 9.53 units then oxygen will be twice as that of carbon.

CO2, so 9.53*2= 19.06 grams of oxygen will combine with 9.53 grams of carbon to form carbon dioxide.

3 0

3 years ago

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

Calculate the amount of heat deposited on the skin of a person burned by 1.00g of steam at 100.0C. Skin temperature is 37C. ΔHva

seraphim [82]

The amount of heat deposited on the skin is 2.26 kJ.

<h3>What is the amount of heat given off by 1.0 g of steam?</h3>

The amount of heat given off by steam is determined using the formula below:

Quantity of heat = mass * latent heat of vaporization.

Moles of steam = 1.0/18

Heat = 1.0/18 * 40.7

Heat deposited = 2.26 kJ

In conclusion, the quantity of heat is determined from the latent heat of vaporization and the moles of steam.

Learn more about heat of vaporization at: brainly.com/question/26306578

#SPJ1

7 0

2 years ago

Calculate the number of moles in a 14.5 gram sample of C4H10.

cricket20 [7]

Moles= mass divided by molar mass
Molar mass= 12.01(4) + 1.01(10)
= 58.14g/mol

Moles=14.5g / 58.14g/mol
=0.249

Therefore there are approx 0.249 moles in a 14.5g sample of C4H10

6 0

3 years ago

Name 5 metals used to make the penny over the years.

Irina18 [472]

These 5 metals are: zinc,copper,tin,bronze,nickel

6 0

3 years ago