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lyudmila [28]
3 years ago
10

Which of the following is a valid mole ratio from the balanced equation 2C3H6 + 902 - 6CO2 + 6H20? (4 points)

Chemistry
1 answer:
erastovalidia [21]3 years ago
6 0

Answer:

6 mol H, 9 mol O2

Explanation:

We can see that in water, there are 6 moles of hydrogen, and for oxygen, there is 9 moles

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A binding protein binds to a ligand l with a kd of 400 nm. what is the concentration of ligand when y is (a) 0.25, (b) 0.6, (c)
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Hey there!:

The fractional saturation y is defined as :

y =  [ L ] / Kd + [ L ]

where :

[ L ] = concentration of binding ligand

Kd = 400 nm

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70 POINTS AND BRAINLIEST ASAPPPP
Crank

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1. Interpret the following equation using moles, molecules, and volumes (assume STP). Compare the mass of the reactants to the m
blsea [12.9K]
1. In this reaction, 2 moles of nitrogen gas reacts with 3 moles of oxygen gas to give 2 moles of N2O3 gas. 2 nitrogen molecules react with 3 oxygen molecules to give 2 N2O3 molecules.  Under STP, one mole of an ideal gas occupies a volume of 22.4 liters. So in this reaction, 44.8 liters of nitrogen gas reacts with 67.2 liters of oxygen gas to give 44.8 liters of N2O3 gas.  The total mass of the reactants (N2 and O2) is the same as the total mass of the product (N2O3). This is called mass balance of a chemical reaction.


2. According to the chemical reaction, 3 moles of chlorine gas produces 2 moles of iron(III) chloride.  So, to produce 1 moles of iron(III) chloride, 3/2 (1.5) moles of chlorine gas is required.  Therefore, to produce 14 moles of iron(III) chloride, 14 x 1.5 = 21 moles of chlorine is needed.
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A rigid tank contains 2.4 kg of helium at determine the volume
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4 years ago
How many grams of CO2 will be produced when 8.50 g of methane react with 15.9 g of O2, according to the following reaction? CH4(
Vedmedyk [2.9K]

Taking into account the reaction stoichiometry, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + 2 O₂  → CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CH₄: 1 mole
  • O₂: 2 moles
  • CO₂:  1 mole
  • H₂O: 2 moles

The molar mass of the compounds is:

  • CH₄: 16 g/mole
  • O₂: 32 g/mole
  • CO₂:  44 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • CH₄: 1 mole ×16 g/mole= 16 grams
  • O₂: 2 moles ×32 g/mole= 64 grams
  • CO₂:  1 mole ×44 g/mole= 44 grams
  • H₂O: 2 moles ×18 g/mole=36 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 16 grams of CH₄ reacts with 64 grams of O₂, 8.50 grams of CH₄ reacts with how much mass of O₂?

mass of O_{2} =\frac{8.50 grams of CH_{4}x64 grams of O_{2} }{16grams of CH_{4}}

mass of O₂= 34 grams

But 34 grams of O₂ are not available, 15.9 grams are available. Since you have less mass than you need to react with 8.50 grams of CH₄, O₂ will be the limiting reagent.

<h3>Mass of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 64 grams of O₂ form 44 grams of CO₂, 15.9 grams of O₂ form how much mass of CO₂?

mass of CO_{2} =\frac{15.9 grams of O_{2}x44 grams of CO_{2} }{64grams of O_{2}}

<u><em>mass of CO₂= 10.93 grams</em></u>

Then, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

6 0
2 years ago
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