CAN SOMEONE PLEASE CHECK MY ANSWER!!!

Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of water is produced from

krek1111 [17]

The given question is incomplete. The complete question is :

Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If 1.31g of water is produced from the reaction of 4.65g of butane and 10.8g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

Answer: 28.0 %

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of butane}=\frac{4.65g}{58g/mol}=0.080moles$

$\text{Moles of oxygen}=\frac{10.8g}{32g/mol}=0.34moles$

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

According to stoichiometry :

13 moles of $O_2$ require 2 moles of butane

Thus 0.34 moles of $O_2$ will require= $\frac{2}{13}\times 0.34=0.052moles$ of butane

Thus $O_2$ is the limiting reagent as it limits the formation of product and butane is the excess reagent.

As 13 moles of $O_2$ give = 10 moles of $H_2O$

Thus 0.34 moles of $O_2$ give = $\frac{10}{13}\times 0.34=0.26moles$ of $H_2O$

Mass of $H_2O=moles\times {\text {Molar mass}}=0.26moles\times 18g/mol=4.68g$

${\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%$

${\text {percentage yield}}=\frac{1.31g}{4.68g}\times 100\%=28.0\%$

The percent yield of water is 28.0 %

6 0

4 years ago

CH3CH2PH2 can be synthesized by an SN2 reaction. Draw the structures of the alkyl chloride and nucleophile that will give this c

Maru [420]

Answer:

The PH2 becomes attached to the alkyl halide with inversion of configuration

Explanation:

SN2 reaction involves the simultaneous attachment of the nucleophile as the leaving group departs. This leads to an inversion of configuration because the sterochemistry around the carbon bearing the leaving group is reversed. The reaction proceed via a crowded five coordinate transition state. Details are shown in the image attached.

6 0

3 years ago

When did the level of oxygen in Earth's atmosphere become high enough to sustain aerobic respiration? When did the level of oxyg

Alexeev081 [22]

Answer:

2.4 billion years ago

Explanation:

The formation of the tertiary atmosphere, with the re-arrangements of the plates, took that influenced the long term evolution of the planet that transferred the carbon dioxide to the large continental stores and the surplus oxygen came into formation about 2.4 billion years ago during the massive oxygenation event that leads to the appearance at the end of the banded iron formations.

The process of photosynthesis also added and reduced the material that reduces the oxidation of the atmosphere and kept ion place till 15% of the Precambrian period had ended.

7 0

3 years ago

Keeping the number of moles constant, what is the final pressure in atm if the temperature was cooled from -91.5°C to -239°C, th

34kurt

Answer:

P2 = 0.935 atm

Explanation:

initial conditions:

∴ V1 = 3.41 L

∴ T1 = - 91.5 °C + 273 = 181.5 K

∴ P1 = 3990 torr * ( atm / 760 torr ) = 5.25 atm

PV = RTn

∴ R = 0.082 atm.L / K.mol

⇒ n = P1V1 / RT1 = ((5.25 atm)*(3.41 L)) / ((0.082 atm.L/K.mol)*(181.5 K))

⇒ n = 1.107 mol

final conditions:

∴ V2 = 3.3 L

∴ T2 = - 239 °C = 34 K

∴ n = 1.107 mol

⇒ P2 = nRT2 / V2

⇒ P2 = ((1.107 mol)*(0.082 atm.L/K.mol)*(34 K)) / 3.3 L

⇒ P2 = 0.935 atm

3 0

3 years ago

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid wat

Mashcka [7]

Answer:

0.26g of NaCl is the maximum mass that could be produced

Explanation:

Based on the reaction:

HCl + NaOH → NaCl + H₂O

Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl

To solve this question we need to find limiting reactant. The moles of limiting reactant = Moles of NaCl produced:

Moles HCl -Molar mass: 36.46g/mol-:

0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl

Moles NaOH -Molar mass: 40g/mol-:

0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH

As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.

The mass of NaCl is:

Mass NaCl -Molar mass: 58.44g/mol-:

0.0045 moles * (58.44g/mol) =

<h3>0.26g of NaCl is the maximum mass that could be produced</h3>

8 0

3 years ago