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stellarik [79]
2 years ago
12

For the following reaction, 4.64 grams of oxygen gas are mixed with excess benzene (C6H6). The reaction yields 3.95 grams of car

bon dioxide. benzene (C6H6) (l) oxygen (g) carbon dioxide (g) water (g) What is the theoretical yield of carbon dioxide
Chemistry
2 answers:
Reil [10]2 years ago
6 0

Answer:

Theoretical yield for CO₂ is 5.10g

Explanation:

Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g)  + 6H₂O(g)

We convert the mass of oxygen to moles:

4.64 g /32 g/mol = 0.145 moles of O₂

Let's find out the 100% yield reaction of CO₂ (theoretical yield)

Ratio is 15:12. So let's make this rule of three:

15 moles of O₂ can produce 12 moles of CO₂

Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles

We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g

uranmaximum [27]2 years ago
3 0

Answer:

The theoretical yield of carbon dioxdide = 5.11 grams

Explanation:

Step 1: Data given

Mass of oxygen gas = 4.64 grams

Molar mass of O2 = 32.0 g/mol

The reaction yields 3.95 grams of carbon dioxide (CO2)

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles oxygen

Moles oxygen = mass oxygen / molar mass oxygen

Moles oxygen = 4.64 grams / 32.0 g/mol

Moles oxygen = 0.145 moles

Step 4: Calculate moles of carbon dioxide (CO2)

For 2 moles C6H6 we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.145 moles O2 we'll have 12/15 * 0.145 = 0.116 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.116 moles * 44.01 g/mol

Mass CO2 = 5.11 grams

The theoretical yield of carbon dioxdide = 5.11 grams

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What is the number of moles of solute in 250 mL of a 0.4 M solution?
mafiozo [28]

Answer:

0,1 mol

Explanation:

We know that the formula of concentration is C= moles of solute/ volume  

0,4 M= moles of solute/ 250 mL

Convert mL to L      250 mL =0,25 L

0,4 M x 0,25 L= moles of solute

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3 0
3 years ago
When sugar glucose, C6H12O6, is burned with air, carbon dioxide and water vapor are produced. Write the balanced equation and ca
Alex Ar [27]
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3 0
3 years ago
II. Ionic Equations
mario62 [17]

Answer:

Complete ionic: \begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Net ionic: \begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

Explanation:

Start by identifying species that exist as ions. In general, such species include:

  • Soluble salts.
  • Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2. These three salts will exist as ions:

  • Each \rm AgNO_3\, (aq) formula unit will exist as one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion.
  • Each \rm CaCl_2 formula unit will exist as one \rm Ca^{2+} ion and two \rm Cl^{-} ions (note the subscript in the formula \rm CaCl_2\!.)
  • Each \rm Ca(NO_3)_2 formula unit will exist as one \rm Ca^{2+} and two \rm {NO_3}^{-} ions.

On the other hand, \rm AgCl is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2 (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each \rm AgNO_3\, (aq) formula unit will exist as only one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion. However, because the coefficient of \rm AgNO_3\, (aq)\! in the original equation is two, \!\rm AgNO_3\, (aq) alone should correspond to two \rm Ag^{+}\! ions and two \rm {NO_3}^{-}\! ions.

Do not rewrite the salt \rm AgCl because it is insoluble.

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of \rm Ca^{2+} and two units of \rm {NO_3}^{-}. Doing so will give:

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}.

Simplify the coefficients:

\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

7 0
2 years ago
What will be the volume occupied by 2.5 moles of nitrogen gas exerting 1.75 atm of pressure at 475K?
Marina86 [1]

Answer:

THE VOLUME OF THE NITROGEN GAS AT 2.5  MOLES , 1.75 ATM AND 475 K IS 55.64 L

Explanation:

Using the ideal gas equation

PV = nRT

P = 1.75 atm

n = 2.5 moles

T = 475 K

R = 0.082 L atm/mol K

V = unknown

Substituting the variables into the equation we have:

V = nRT / P

V = 2.5 * 0.082 * 475 / 1.75

V = 97.375 / 1.75

V = 55.64 L

The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L

6 0
2 years ago
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