The reaction can be described as follows: CO + 2H2 = CH3OH. Given the specified quantities of the reactants, we will identify the limiting reactant and compute the remaining excess amount. Calculating, 1.50 x 10^-6 g CO converts to 5.36 x 10^-8 mol CO, while 6.80 x 10^-6 g H2 equals 3.37 x 10^-6 mol H2. Thus, CO is fully consumed in the reaction, leaving 3.37 x 10^-6 - 5.36 x 10^-8 = 3.32 x 10^-6 moles of gas.
The visual representation is displayed in the following image.
For calculations, consider 100 grams of the compound:
ω(Cl) = 85.5% ÷ 100%.
ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.
m(Cl) = 0.855 · 100 g.
m(Cl) = 85.5 g; this represents the mass of chlorine.
m(C) = 100 g - 85.5 g.
m(C) = 14.5 g; indicating the mass of carbon.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 85.5 g ÷ 35.45 g/mol.
n(Cl) = 2.41 mol; this is the quantity of chlorine.
n(C) = 14.5 g ÷ 12 g/mol.
n(C) = 1.21 mol; this is the quantity of carbon.
n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.
The compound in question is identified as dichlorocarbene CCl₂.
1) The chemical formula for propane is CH₃-CH₂-CH₃.
Propane is classified as a three-carbon alkane (acyclic saturated <span>hydrocarbon).
</span>2) The chemical formula for propanal is CH₃-CH₂-CH=O.
Propanal <span> is a </span>saturated<span> three-carbon </span>aldehyde (consists of<span> a </span>carbonyl<span> center).
3) </span>The chemical formula for propanol is CH₃-CH₂-CH₂-OH.
1-propanol <span> is a </span><span>primary alcohol.
4) </span>The chemical formula for propanone is (CH₃)₂-C=O.
Propanone, also known as acetone, is <span>the simplest and smallest</span> ketone.
The correct answer is the fifth option. Energy transfers from the fire to the pot, subsequently to the water, and then to the peas.
