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Eva8 [605]
4 years ago
11

Examine the equation.

Chemistry
2 answers:
jenyasd209 [6]4 years ago
7 0

Answer:

Coefficient

Explanation:

We can explain this discarting options too. Now, let's write again the chemical equation;

C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)

In a chemical equation, we have two sides. The left side we have the reactants of the reaction, these can be elements or compounds that reacts between each other, to produce another compound or substance. This compounds or elements are on the right side of the reaction. This side we have the products of the reaction of reactants.

Now, as in every situation of life, we need a balance of reactants and products, so, the only way to balance a chemical equation is by putting numbers behind every compound of the reactant and/or products to equal the number of atoms on each side. So if in the reactants we have 3 atoms of C, in the products we need to have the same 3 atoms.

Based on the above explanation, we can say that the number 4 is behind the H2O. This compound is on the side of products, but the number only, do not represent a product, so this option can be discarted.

A reactant cannot be either, cause it's not on the side of reactant.

Chemical formula, could be, if the number were as subscript, but it's not the case, and the chemical formula is composed of symbols.

Finally, as the number that we put behind the compounds work to balance the equation, the only function that these numbers have here, is to be coefficients of the reactions, and furthermore, it will work as mole.

So the correct option is the third one, Coefficient.

azamat4 years ago
3 0
4 is a coefficient of the formula
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CH3OH can be synthesized by the following reaction.
puteri [66]

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A: There is 43.12 Liter of H2 needed

B: There is 21.56 liter of CO needed

Explanation:

Step 1: Data given

CO(g)+2H2(g)?CH3OH(g)

For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

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Step 2: What volume of H2 gas (in L), measured at 754mmHg and 90?C, is required to synthesize 23.0g CH3OH?

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Temperature = 90°C = 363 Kelvin

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Step 3: Calculate moles of CH3OH

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Moles CH3OH = 23.0 grams / 32.04 g/mol

Moles CH3OH = 0.718 moles

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For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

For 0.718 moles CH3OH produced, we have 2*0.718 moles =1.436 moles of H2 and 0.718 moles of CO

Step 5: Calculate volume of H2

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 1.436 moles H2

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (1.436*0.08206*363)/0.992

V = 43.12 L

Step 6: Calculate volume of CO

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 0.718  moles CO

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (0.718*0.08206*363)/0.992

V = 21.56 L

A: There is 43.12 Liter of H2 needed

B: There is 21.56 liter of CO needed

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3 years ago
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