Methane gas can be made from carbon (II) oxide gas according to the equation 2CO(g)...
Methane gas can be made from carbon (II) oxide gas according to the equation 2CO(g) + 2H\(_2\)(g) \(\rightarrow\) CH\(_4\)(g) + CO\(_2\)(g). Calculate the mass of CO required to produce 8.75 \(\times\) 10\(^{25}\) molecules of CH\(_4\)? [At masses: C = 12.011, H = 1.008, O = 15.999, Avogadro's No.: 6.022 \(\times\) 10\(^{23}\) molecules /mole.]
- A) 8140g
- B) 4070g
- C) 1600g
- D) 32.00g
Correct Answer: A) 8140g
Explanation
First, let's find the number of moles of CH\(_4\) produced in the reaction. We can do this by using Avogadro's Number, which tells us the number of particles (atoms, molecules, ions, etc.) in one mole of a substance.
To find the number of moles of CH\(_4\), divide the given number of molecules by Avogadro's Number:
\[\text{moles of CH}_4 = \frac{8.75 \times 10^{25}\text{ molecules}}{6.022 \times 10^{23}\text{ molecules/mole}} = 145\text{ moles}\]Now, we can use the stoichiometry of the balanced chemical equation to find the number of moles of CO required. As per the equation, 2 moles of CO are required to produce 1 mole of CH\(_4\):
\[\text{moles of CO} = 2 \times \text{moles of CH}_4 = 2 \times 145\text{ moles} = 290\text{ moles}\]Next, we will find the molar mass of CO:
\[\text{molar mass of CO} = \text{molar mass of C} + \text{molar mass of O} = 12.011\text{ g/mol} + 15.999\text{ g/mol} = 28.01\text{ g/mol}\]Finally, we can calculate the mass of CO required by multiplying the number of moles of CO by its molar mass:
\[\text{mass of CO} = \text{moles of CO} \times \text{molar mass of CO} = 290\text{ moles} \times 28.01\text{ g/mol} = 8140\text{ g}\]So, the mass of CO required to produce 8.75 \(\times\) 10\(^{25}\) molecules of CH\(_4\) is8140g, which isOption A.