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lapo4ka [179]
2 years ago
10

The complete combustion (reaction with oxygen) of liquid octane (C8H18) a component typical of the hydrocarbons in gasoline, pro

duces carbon dioxide gas and water vapor. What is the coefficient of octane in the balanced equation for the reaction? (Balance the equation with the smallest possible whole number coefficients.) A.2 B.15 C.4 D3 E. 24 F. 20 G. 25
Chemistry
1 answer:
lozanna [386]2 years ago
4 0

Answer:

The answer to your question is letter A. 2

Explanation:

Data

Octane = C₈H₁₈

Oxygen = O₂

Carbon dioxide = CO₂

Water = H₂O

Reaction

                         C₈H₁₈  +   O₂   ⇒   CO₂   +   H₂O

                  Reactants         Elements          Products

                          8                     C                         1

                         18                     H                        2

                          2                     O                        3

This reaction is unbalanced

                          C₈H₁₈  +  25/2O₂   ⇒   8CO₂   +   9H₂O

                  Reactants         Elements          Products

                          8                     C                         8

                         18                     H                        18

                         25                     O                       25

Multiply the reaction by 2

                    2C₈H₁₈  +   25O₂   ⇒   16CO₂   +   18H₂O

                  Reactants         Elements          Products

                         16                     C                        16

                         36                     H                       36

                         50                     O                       50

Now, the reaction is balanced

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Answer:

3.43 moles

Explanation:

To convert moles into grams you must take the number of grams and divide it by the atomic mass of the compound.

in this case, grams is 151 and N2O has an atomic mass of 44.013

so your equation would look like

151/44.013=3.430804535

your answer would be 3.43 moles

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Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of c
Anna [14]

The question is incomplete. The complete question is

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed

Answer:

a)84.91g

b)8.20g

c)316.4g

d)616.73g

Explanation:

The equation of the reaction:

2K3PO4(aq) + 3Ca(NO3)2 (aq)-------> 6KNO3(aq) + Ca3(PO4)2(s)

Molar mass of potassium phosphate= 212.27 g/mol

Amount of potassium phosphate= 500/212.27= 2.4 moles

Molar mass of calcium nitrate= 164.088 g/mol

Amount of calcium nitrate= 500/164.088=3.05moles

a) amount of potassium phosphate reacted according to reaction equation= 2 moles

Amount of potassium phosphate remaining= 2.4-2=0.4 moles

Mass of potassium phosphate remaining= 0.4×212.27=84.91g

b) Amount of calcium nitrate reacted according to reaction equation=3

Amount of calcium nitrate remaining=3.05-3= 0.05

Mass of calcium nitrate remaining= 0.05×164.088= 8.20g

c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.

From the reaction equation,

3 moles of calcium nitrate yields 1 mole of calcium phosphate

3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate

Molar Mass of calcium phosphate= 310.18 g/mol

Mass of calcium phosphate produced= 1.02×310.18= 316.4g

d)

3 moles of calcium nitrate yields 6 moles of potassium nitrate

3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate

Molar mass of potassium nitrate = 101.1032 g/mol

Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g

6 0
3 years ago
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