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4 years ago

Consider the following reaction:

Chemistry

2 answers:

earnstyle [38]4 years ago

8 0

Answer : The amount of heat transferred in the reaction is 213.4 kJ

Explanation :

Mass of $CH_3OH$ = 27.0 g

Molar mass of $CH_3OH$ = 32 g/mole

Enthalpy of reaction = +252.8 kJ

First we have to calculate the moles of $CH_3OH$ .

$\text{ Moles of }CH_3OH=\frac{\text{ Mass of }CH_3OH}{\text{ Molar mass of }CH_3OH}=\frac{27.0g}{32g/mole}=0.844moles$

Now we have to calculate the amount of heat transferred in the reaction.

The given balanced chemical reaction is:

$2CH_3OH(g)\rightarrow 2CH_4(g)+O_2(g)$

From the reaction, we conclude that

As, 2 moles of $CH_3OH$ transfer heat = 252.8 kJ

So, 0.844 moles of $CH_3OH$ transfer heat = 0.844 × (252.8 kJ)

= 213.4 kJ

Therefore, the amount of heat transferred in the reaction is 213.4 kJ

Anni [7]4 years ago

6 0

<span>Consider the following reaction:
2CH3OH(g)→2CH4(g)+O2(g)ΔH=+252.8kJ
Calculate the amount of heat transferred when 27.0g of CH3OH(g) is decomposed by this reaction at constant pressure.

27.0 g </span>CH3OH(g) (1 mol/ 32.05g) = 0.84 mol CH3OH(g)
ΔH =+252.8kJ = Q = 0.84 mol (252.8kJ) = 213 kJ

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3 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

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4 years ago

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$