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3 years ago

How many moles of CO2 will be produced from 97.0 g of C3 Hy, assuming O2 is available in excess?

Chemistry

1 answer:

xeze [42]3 years ago

3 0

Answer:

291 / (36 + y)

Explanation:

Regardless of the value of the y subscript, the number of moles of CO₂ that can be produced from 1 mol of $C_3H_y$ is 3 moles: that is so because every molecule of $C_3H_y$ contains 3 atoms of C and every atom of C produces one molecule of CO₂.

Then, you can calculate the number of moles of $C_3H_y$ in 97.0g using the molar mass of the compound:

1. Molar mass of $C_3H_y$ :

3 × 12g/mol + y × 1 g/mol = (36 + y)g/mol

2. Number of moles = mass in grams / molar mass

Number of moles of $C_3H_y$ =

= 97.0 g / (36 + y)g/mol = 97.0/(36+y) mol

3. Number of moles of CO₂

As stated above, the number of moles of CO₂ is 3 times the number of moles of $C_3H_y$ :

Number of moles of CO₂ = 3 × 97.0 / (36 + y) mol = 291 / (36 + y)

For instance, imagine the compound is C₃H₈. How many moles of CO₂ will be produced from 97.0 g of C₃H₈?

You can replace 8 for y:

291 / (36 + 8) mol = 291.0 / (44) mol = 6.61 mol

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3 years ago

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0

2 years ago

For the following reaction, the equilibrium constant Kc is 2.0 at a certain temperature. The reaction is endothermic. What do yo

jeyben [28]

Answer:

The correct answer is : 'the concatenation of NO will increase'.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in temperature occurs.

$2NOBr(g)\rightleftharpoons 2NO(g) + Br_2(g)$

As, this is an endothermic reaction, increasing temperature will add more heat to the system which move equilibrium in the forward reaction with decrease in temperature. Hence, the equilibrium will shift in the right direction.

So, the concatenation of NO will increase.

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3 years ago