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3 years ago

8

Given the following equation: > 8 CO2 10 H20 2 С4Н10 + 13 02 How many grams of C4H10 are needed to react with 35.1 grams of O

2?

1 answer:

PilotLPTM [1.2K]3 years ago

4 0

<u>Answer:</u> The mass of butane reacting with oxygen gas is 9.76 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For oxygen</u>

Given mass of oxygen gas = 35.1 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{35.1g}{32g/mol}=1.096mol$

For the given chemical reaction:

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

By stoichiometry of the reaction:

13 moles of oxygen gas is reacting with 2 moles of butane.

So, 1.096 moles of oxygen gas will react with = $\frac{2}{13}\times 1.096=0.168moles$ of butane.

Now, calculating the mass of butane from equation 1, we get:

Molar mass of butane = 58.12 g/mol

Moles of butane = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of butane}}{58.12g/mol}\\\\\text{Mass of butane}=9.76g$

Hence, the mass of butane reacting with oxygen gas is 9.76 grams.

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34.5 X 10^-11 grams of lead

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3 years ago

Please help with #2 and #3

Vaselesa [24]

Answer:

2. $V_2=17L$

3. $V=82.9L$

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

$\frac{V_2}{T_2}=\frac{V_1}{T_1}$

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

$V_2=\frac{V_1T_2}{T_1}$

Therefore, we plug in the given data to obtain:

$V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L$

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

$V=\frac{nRT}{P}$

Therefore, we plug in the data to obtain:

$V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L$

Best regards!

6 0

3 years ago

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Tom [10]

Answer:

Explanation:

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5 0

4 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

We drop a cube of ice into a glass of water. The mass of the cube of ice is 33.1 g and its initial temperature is −10.2∘C. The m

Nastasia [14]

Answer: The final temperature (T) will be;

16.23°c

Explanation: To find the final temperature of any mixture of substance with an initial temperature use the formula;

M1c(T-T1) + M2c(T-T2)=0

c is the specific heat capacity of the two different substance, but because ice and water are the same, we assume c to be 1

T= final temperature of the mixture

T1= initial temperature of the ice= -10.2°c

T2= initial temperature of the water= 19.7°c

M1= mass of ice= 33.1g

M2= mass of water= 251g

Using the formula above

33.1(T-(-10.2)) + 251(T-19.7)=0

Solving out the bracket

33.1T + 337.62 + 251T - 4944.7 = 0

Collecting like terms to both side of the equation and solving

33.1T + 251T = 4944.7 - 33.62

284.1T = 4607.08

T = 4607.08÷284.1 = 16.23°c

6 0

3 years ago