Answer:
35.6 liters at STP
Explanation:
The molar mass of carbon dioxide is about 44.01 g/mol. The volume of a mole of ideal gas at STP is 22.4 L, so the volume of 70.0 g will be ...
(70.0g)/(44.01 g/mol)·(22.4 L/mol) ≈ 35.6 L
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
Answer:
Explanation:
Sodium chloride is ionic compound. It is formed by the transfer of electron from one atom to the atom of another element.
Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.
The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion. Both atoms are bonded together electrostatic attraction occur between anion and cations.
Sodium atom have one valance electron by losing this one valance electron sodium atom get the complete octet. Chlorine atom has seven valance electrons and needed to lose seven valance electrons or to get one electron and thus complete the octet. It is very easy for chlorine atom to get one electrons instead of losing all seven electron. Thus when it react with sodium it gain the valance electron of sodium and form ionic compound.
That's why only one atom of sodium combine with one atom chlorine.
Multiple by 2 because it was give u right answer
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
