Answer:
Explanation: n=m/M(molar mass)
n=24.3 grams/(16+2x1.008)grams/moles(molar mass of H2O)
n=24.3grams/18.016grams/moles
n=1.35moles
D.
The products will have at least 2 Na atoms and 1 O atom.
<h3>Further explanation</h3>
If we refer to the law of mass conservation, which states that
<em>In a closed system, the masses before and after the reaction are the same
</em>
then the number of atoms in the reactance will be the same as the number of atoms in the product
In this problem it is known that Na₂O is one of the reactants so that the product of Na atoms and O atoms will at least equal the number of atoms in the bond, namely 2 Na and 1 O
Like an example of this Na₂O reaction:
Na₂O + H₂O ⇒ 2 NaOH
Na : left =2, right = 2
O : left=2, right = 2
Formic acid is the simplest carboxylic acid with a structure of HCOOH and has a pka of 3.75. The pka refers to the acidity of the molecule, which in this example refers to the molecules ability to give up the proton of the O-H. A decrease in the pka value corresponds to an increase in acidity, or an increase in the ability to give up a proton. When an acid gives up a proton, the remaining anionic species (in this case HCOO-) is called the conjugate base, and an increase in the stability of the conjugate base corresponds to an increase in acidity.
The pka of a carboxylic can be affected greatly by the presence of various functional groups within its structure. An example of an inductive effect changing the pka can be shown with trichloroacetic acid, Cl3CCOOH. This molecule has a pka of 0.7. The decrease in pka relative to formic acid is due to the presence of the Cl3C- group, and more specifically the presence of the chlorine atoms. The electronegative chlorine atoms are able to withdraw the electron density away from the oxygen atoms and towards themselves, thus helping to stabilize the negative charge and stabilize the conjugate base. This results in an increase in acidity and decrease in pka.
The same Cl3CCOOH example can be used to explain how dipoles can effect the acidity of carboxylic acids. Compared to standard acetic acid, H3CCOOH with a pka of 4.76, trichloroacetic acid is much more acidic. The difference between these structures is the presence of C-Cl bonds in place of C-H bonds. A C-Cl bond is much more polar than a C-H bond, due the large electronegativity of the chlorine atom. This results in a carbon with a partial positive charge and a chlorine with a partial negative charge. In the conjugate base of the acid, where the molecule has a negative charge localized on the oxygen atoms, the dipole moment of the C-Cl bond is oriented such that the partial positive charge is on the carbon that is adjacent to the oxygen atoms containing the negative charge. Therefore, the electrostatic attraction between the positive end of the C-Cl dipole and the negative charge of the anionic oxygen helps to stabilize the entire species. This level of stabilization is not present in acetic acid where there are C-H bonds instead of C-Cl bonds since the C-H bonds do not have a large dipole moment.
To understand how resonance can affect the pka of a species, we can simply compare the pka of a simple alcohol such as methanol, CH3OH, and formic acid, HCOOH. The pka of methanol is 16, suggesting that is is a very weak acid. Once methanol gives up that proton to become the conjugate base CH3O-, the charge cannot be stabilized in any way and is simply localized on the oxygen atom. However, with a carboxylic acid, the conjugate base, HCOO-, can stabilize the negative charge. The lone pair electrons containing the charge on the oxygen atom are able to migrate to the other oxygen atom of the carboxylic acid. The negative charge can now be shared between the two electronegative oxygen atoms, thus stabilizing the charge and decreasing the pka.
The heat of reaction (i.e. combustion) of butane (
) when reacted with oxygen (
) is -2658 kJ/mol butane, and the chemical reaction is given by:
+
--->
+
The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.
Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.
Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane
Mass of butane = 0.5643 moles butane * 58 grams/mol butane
Mass of butane =
32.73 grams butane
The mass of carbon dioxide (
) can be determined by multiplying the moles of butane (
) with the mole ratio of (
) produced to the (
) reacted, and then with the molar mass of (
), which is 44 grams/mole.
Mass of carbon dioxide produced = 0.5643 moles butane * [4 moles
/ 1 mole
] * 44 grams/mole
Mass of carbon dioxide produced
= 99.32 grams Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.
