$0.8 \text{ dm}^3$ of CO, $0.6 \text{ dm}^3$ of $CO_2$
Reaction Analysis: CO2 with Hot Coke
The reaction between carbon dioxide ($CO_2$) gas and hot coke (carbon, $C$) produces carbon monoxide (CO) gas. The balanced chemical equation is:
$CO_2(g) + C(s) \rightarrow 2CO(g)$
This equation shows that 1 volume of $CO_2$ reacts to produce 2 volumes of CO. This results in a net increase in the volume of gaseous products compared to the reactant consumed.
Volume Change Calculation
Let the initial volume of $CO_2$ gas be $V_{CO_2, initial} = 1 \text{ dm}^3$. Let $x$ represent the volume of $CO_2$ that reacts completely.
- Volume of $CO_2$ consumed = $x \text{ dm}^3$.
- Volume of CO produced = $2x \text{ dm}^3$ (based on the stoichiometry).
- Volume of $CO_2$ remaining unreacted = $V_{CO_2, initial} - x = (1 - x) \text{ dm}^3$.
The total final volume of the gaseous mixture ($V_{final}$) is the sum of the unreacted $CO_2$ and the produced CO:
$V_{final} = (\text{Volume of } CO_2 \text{ remaining}) + (\text{Volume of } CO \text{ produced})$
$V_{final} = (1 - x) + 2x$
$V_{final} = 1 + x$
The problem states that the final volume after the complete reaction is $1.4 \text{ dm}^3$. Therefore:
$1 + x = 1.4 \text{ dm}^3$
Solving for $x$ gives:
$x = 1.4 - 1 = 0.4 \text{ dm}^3$
Final Gas Composition
Now, we calculate the volume of each gas present in the final mixture:
- Volume of CO produced = $2x = 2 \times 0.4 = 0.8 \text{ dm}^3$.
- Volume of $CO_2$ remaining = $1 - x = 1 - 0.4 = 0.6 \text{ dm}^3$.
Thus, the composition of the gaseous mixture at STP is $0.8 \text{ dm}^3$ of CO and $0.6 \text{ dm}^3$ of $CO_2$.
Similar Questions
- In a qualitative analysis, $Bi^{3+}$ is detected by appearance of precipitate of $BiO(OH)(s)$. Calculate pH when the following equilibrium exists at 298 K :
$$BiO(OH) (s) \rightleftharpoons BiO^+ (aq) + OH^- (aq)$$
$K = 4 \times 10^{-10}$
(Given : $\log 2 = 0.3010$) Match List I with List II :
List I (Quantum Numbers )
$n$, $l$
List II (Orbital) A. 2, 1 I. 3d B. 4, 0 II. 2p C. 5, 3 III. 4s D. 3, 2 IV. 5f Choose the correct answer from the options given below :
- A solution of copper sulphate is electrolysed for $10 \text{ minutes}$ with a current of $1.5 \text{ amperes}$. The mass of copper deposited at cathode is :
(Given : Molar mass of Cu = $63 \text{ g mol}^{-1}$, $1 \text{ F} = 96487 \text{ C mol}^{-1}$) - Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, T (K).
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is :
(Given : $k \text{ in s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$) - Match List I with List II :
List I (Order of reaction) List II (Unit of rate constant) A. Zero order I. $mol^{-1} L s^{-1}$ B. First order II. $mol^{-2} L^2 s^{-1}$ C. Second order III. $s^{-1}$ D. Third order IV. $mol L^{-1} s^{-1}$
Choose the correct answer from the options given below : At 298 K, a certain buffer solution contains equal concentrations of $X^{-}$ and $HX$. $K_b$ for $X^-$ is $10^{-10}$. What is the pH of this buffer solution ?
- At a certain temperature T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is :
Consider the following reaction :
$2A (g) + B (g) \rightarrow 2D (g)$
$\Delta U^\ominus = -10 \text{ kJ mol}^{-1}$ and $\Delta S^\ominus = -44 \text{ J K}^{-1}$ at 298 K.
Identify the correct option with $\Delta G^\ominus$ for the reaction and spontaneity of the reaction at 298 K. (Given : $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$)- A bulb is rated at 150 watt, converting 8% energy into light. If energy of one photon is $4.42 \times 10^{-19} \text{ J}$, how many photons are emitted by the bulb per second ?
- Given below are certain reactions. Identify the reaction for which $K_p \neq K_c$.
Important Questions from Physical Chemistry
- X and Y are the number of electrons involved, respectively during the oxidation of $\text{I}^-$ to $\text{I}_2$ and $\text{S}^{2-}$ to S by acidified $\text{K}_2\text{Cr}_2\text{O}_7$. The value of X + Y is ______.
- Consider two Group IV metal ions $\text{X}^{2+}$ and $\text{Y}^{2+}$.
A solution containing 0.01 M $\text{X}^{2+}$ and 0.01 M $\text{Y}^{2+}$ is saturated with $\text{H}_2\text{S}$. The pH at which the metal sulphide YS will form as a precipitate is ______. (Nearest integer)
(Given: $\text{K}_{sp}(\text{XS}) = 1 \times 10^{-22}$ at $25^\circ\text{C}$, $\text{K}_{sp}(\text{YS}) = 4 \times 10^{-16}$ at $25^\circ\text{C}$, $[\text{H}_2\text{S}] = 0.1\text{M}$ in solution, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$) - Electricity is passed through an acidic solution of $\text{Cu}^{2+}$ till all the $\text{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer)
[Given:
$\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \quad \text{E}^\circ_{\text{red}} = +0.34 \text{ V}$
$\text{O}_2\text{(g)} + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O} \quad \text{E}^\circ_{\text{red}} = +1.23 \text{ V}$
Molar mass of Cu = $63.54 \text{ g mol}^{-1}$
Molar mass of $\text{O}_2$ = $32 \text{ g mol}^{-1}$
Faraday Constant = $96500 \text{ C mol}^{-1}$
Molar volume at STP = 22.4 L] - Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and $15 \text{ kN m}^{-2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?
- At 298 K, the mole percentage of $\text{N}_2\text{(g)}$ in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of $\text{N}_2\text{(g)}$ in water at 298 K?
($K_H \text{ for } \text{N}_2 \text{ is } 6.5 \times 10^7 \text{ mm Hg}$)
