Question

Find the number of molecule of CNS 2/2 if Y to produce 0.25 moles of and no and find the volume of O2 produced on STP also find the number of atoms present in reaction at the given condition

Answer 1

Number of molecules of CNS 2/2 = 7.5275 × 10^22 molecules, Volume of O2 required at STP = 14 L, Number of atoms in products = 1.2044 × 10^24 atoms

Answer 2

The problem statement is poorly phrased and contains ambiguities and likely typos. We will proceed by making the most plausible assumptions to make the problem solvable, aligning with typical stoichiometry questions.

Assumptions:

1. "CNS 2/2": This is interpreted as the chemical formula $C_2N_2S_2$ (thiocyanogen). The "Y" refers to this compound.
2. "0.25 moles of and no": This is interpreted as "0.25 moles of NO (nitric oxide)". NO is a product.
3. "volume of O2 produced on STP": Given that $C_2N_2S_2$ contains C and S, a reaction involving $O_2$ is likely a combustion/oxidation. In such reactions, $O_2$ is typically a reactant. Therefore, "produced" is assumed to be a typo for "required".
4. "number of atoms present in reaction at the given condition": Similar to the provided example, this is interpreted as the total number of atoms in the products formed under the given conditions.

Step 1: Write and Balance the Chemical Equation Assuming $C_2N_2S_2$ reacts with $O_2$ to produce $NO$, $CO_2$, and $SO_2$: $C_2N_2S_2 + O_2 \rightarrow NO + CO_2 + SO_2$

Balance the equation:

• Balance C: 2 C on left, so $2CO_2$ on right.
• Balance N: 2 N on left, so $2NO$ on right.
• Balance S: 2 S on left, so $2SO_2$ on right.
• Now, balance O: Reactant side: $x O_2$ Product side: $2 \text{ (from NO)} + 2 \times 2 \text{ (from CO2)} + 2 \times 2 \text{ (from SO2)} = 2 + 4 + 4 = 10$ oxygen atoms. So, $5O_2$ on the reactant side.

The balanced chemical equation is: $C_2N_2S_2 (s) + 5O_2 (g) \rightarrow 2NO (g) + 2CO_2 (g) + 2SO_2 (g)$

Step 2: Calculate the Number of Molecules of $C_2N_2S_2$ From the balanced equation, 1 mole of $C_2N_2S_2$ produces 2 moles of NO. Given that 0.25 moles of NO are produced. Moles of $C_2N_2S_2$ required = $0.25 \text{ mol NO} \times \frac{1 \text{ mol } C_2N_2S_2}{2 \text{ mol NO}} = 0.125 \text{ mol } C_2N_2S_2$.

Number of molecules = Moles $\times$ Avogadro's Number ($N_A = 6.022 \times 10^{23} \text{ molecules/mol}$) Number of molecules of $C_2N_2S_2 = 0.125 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}$ Number of molecules of $C_2N_2S_2 = 7.5275 \times 10^{22}$ molecules.

Step 3: Calculate the Volume of $O_2$ Required at STP From the balanced equation, 5 moles of $O_2$ are required to produce 2 moles of NO. Moles of $O_2$ required = $0.25 \text{ mol NO} \times \frac{5 \text{ mol } O_2}{2 \text{ mol NO}} = 0.625 \text{ mol } O_2$.

At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L. Volume of $O_2$ required = Moles of $O_2 \times$ Molar volume at STP Volume of $O_2$ required = $0.625 \text{ mol} \times 22.4 \text{ L/mol} = 14 \text{ L}$.

Step 4: Calculate the Total Number of Atoms Present in the Products The products are NO, $CO_2$, and $SO_2$. We know 0.25 moles of NO are produced. From the balanced equation: Moles of $CO_2$ produced = $0.25 \text{ mol NO} \times \frac{2 \text{ mol } CO_2}{2 \text{ mol NO}} = 0.25 \text{ mol } CO_2$. Moles of $SO_2$ produced = $0.25 \text{ mol NO} \times \frac{2 \text{ mol } SO_2}{2 \text{ mol NO}} = 0.25 \text{ mol } SO_2$.

Calculate the number of atoms per molecule for each product:

• NO: 1 N atom + 1 O atom = 2 atoms/molecule
• $CO_2$: 1 C atom + 2 O atoms = 3 atoms/molecule
• $SO_2$: 1 S atom + 2 O atoms = 3 atoms/molecule

Total moles of atoms in products: Atoms from NO = $0.25 \text{ mol NO} \times 2 \text{ atoms/molecule} = 0.5 \text{ mol atoms}$. Atoms from $CO_2 = 0.25 \text{ mol } CO_2 \times 3 \text{ atoms/molecule} = 0.75 \text{ mol atoms}$. Atoms from $SO_2 = 0.25 \text{ mol } SO_2 \times 3 \text{ atoms/molecule} = 0.75 \text{ mol atoms}$.

Total moles of atoms in products = $0.5 + 0.75 + 0.75 = 2 \text{ mol atoms}$.

Total number of atoms in products = Total moles of atoms $\times N_A$ Total number of atoms in products = $2 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol}$ Total number of atoms in products = $1.2044 \times 10^{24}$ atoms.

Summary of Results:

• Number of molecules of $C_2N_2S_2$: $7.5275 \times 10^{22}$ molecules
• Volume of $O_2$ required at STP: 14 L
• Total number of atoms present in the products: $1.2044 \times 10^{24}$ atoms