1. How many moles of Al(CN)3 are in 199 g of the compound?
2. 96.0 mol of P4O10 contains how many moles of P?
3. If 4.71 moles of C5H12 reacts with excess O2, how many moles of CO2 will be produced in the following combustion reaction? C5H12+8O2 -> 6H20+5CO2
4. How many grams of O2(g) are needed to completely burn 51.0 g of C3H8(g)?
C3H8+5O2 -> 3CO2+4H2O
5. How many grams of sodium iodide, NaI, must be used to produce 69.9 g of iodine, I2? 2NaI+Cl2 -> I2+2NaCl
6. What is the average atomic mass of titanium on that planet?
Isotope: 46Ti
Abudance: 70.900%
Mass(amu): 45.95263
Isotope: 48Ti
Abudance: 10.000%
Mass(amu): 47.94795
Isotope: 50Ti
Abundance: 19.100%
Mass(amu): 49.94479
1. To find the number of moles of Al(CN)3 in 199 g of the compound, you need to divide the mass of the compound by its molar mass.
The molar mass of Al(CN)3 can be calculated by summing the atomic masses of its constituent elements:
Aluminum (Al) has a molar mass of 26.98 g/mol.
Carbon (C) has a molar mass of 12.01 g/mol.
Nitrogen (N) has a molar mass of 14.01 g/mol.
Therefore, the total molar mass of Al(CN)3 is:
Al: (26.98 g/mol) x 1 = 26.98 g/mol
C: (12.01 g/mol) x 1 = 12.01 g/mol
N: (14.01 g/mol) x 3 = 42.03 g/mol
Adding these together: 26.98 + 12.01 + 42.03 = 81.02 g/mol
To find the number of moles of Al(CN)3 in 199 g, divide the mass by the molar mass:
Number of moles = Mass / Molar mass
Number of moles = 199 g / 81.02 g/mol = 2.4599 mol (rounded to 4 decimal places)
Therefore, there are approximately 2.4599 moles of Al(CN)3 in 199 g of the compound.
2. To find the number of moles of P in 96.0 mol of P4O10, you need to use the balanced chemical equation for the reaction that involves P4O10.
The balanced equation is:
P4O10 + 6P → 4P2O5
According to the equation, there are 6 moles of P for every 1 mole of P4O10.
Therefore, the number of moles of P in 96.0 mol of P4O10 is:
Number of moles of P = 96.0 mol P4O10 × 6 mol P / 1 mol P4O10 = 576 mol
Therefore, there are 576 moles of P in 96.0 mol of P4O10.
3. In the combustion reaction C5H12 + 8O2 → 6H2O + 5CO2, the stoichiometric ratio between C5H12 and CO2 is 1:5.
That means for every mole of C5H12, 5 moles of CO2 are produced.
Given that there are 4.71 moles of C5H12, the number of moles of CO2 produced is:
Number of moles of CO2 = 4.71 mol C5H12 × 5 mol CO2 / 1 mol C5H12 = 23.55 mol
Therefore, 4.71 moles of C5H12 will produce 23.55 moles of CO2 in the combustion reaction.
4. In the balanced chemical equation C3H8 + 5O2 → 3CO2 + 4H2O, the stoichiometric ratio between C3H8 and O2 is 1:5.
Therefore, for every mole of C3H8, 5 moles of O2 are required.
To find the number of moles of O2, you can use the molar mass of C3H8 and the given mass of C3H8.
The molar mass of C3H8 can be calculated by summing the atomic masses of its constituent elements:
C: (12.01 g/mol) x 3 = 36.03 g/mol
H: (1.01 g/mol) x 8 = 8.08 g/mol
Adding these together: 36.03 + 8.08 = 44.11 g/mol
To find the number of moles of C3H8, divide the given mass by its molar mass:
Number of moles of C3H8 = Mass / Molar mass
Number of moles of C3H8 = 51.0 g / 44.11 g/mol = 1.1575 mol (rounded to 4 decimal places)
Since the stoichiometric ratio is 1:5, this means that 5 times the number of moles of O2 is required:
Number of moles of O2 = 1.1575 mol C3H8 × 5 mol O2 / 1 mol C3H8 = 5.7876 mol (rounded to 4 decimal places)
Therefore, approximately 5.7876 moles of O2 are needed to completely burn 51.0 g of C3H8.
5. In the balanced chemical equation 2NaI + Cl2 → I2 + 2NaCl, the stoichiometric ratio between NaI and I2 is 2:1.
That means for every 2 moles of NaI, 1 mole of I2 is produced.
To find the number of moles of NaI needed to produce 69.9 g of I2, you need to use the molar mass of I2 and the given mass of I2.
The molar mass of I2 is the sum of the atomic masses of iodine:
I: (126.90 g/mol) x 2 = 253.80 g/mol
To find the number of moles of I2, divide the given mass by its molar mass:
Number of moles of I2 = Mass / Molar mass
Number of moles of I2 = 69.9 g / 253.80 g/mol = 0.2757 mol (rounded to 4 decimal places)
Since the stoichiometric ratio is 2:1, this means that half the number of moles of NaI is required:
Number of moles of NaI = 0.2757 mol I2 × 2 mol NaI / 1 mol I2 = 0.5514 mol (rounded to 4 decimal places)
Therefore, approximately 0.5514 moles of NaI must be used to produce 69.9 g of I2.
6. To find the average atomic mass of titanium on that planet, you need to calculate the weighted average of the atomic masses of the three isotopes.
The average atomic mass of a element is calculated by multiplying the mass of each isotope by its abundance and summing these values for all isotopes.
For isotope 46Ti:
Mass (amu) = 45.95263 amu
Abundance = 70.900%
For isotope 48Ti:
Mass (amu) = 47.94795 amu
Abundance = 10.000%
For isotope 50Ti:
Mass (amu) = 49.94479 amu
Abundance = 19.100%
To calculate the average atomic mass, multiply the mass of each isotope by its abundance and sum these values:
Average Atomic Mass = (Mass of Isotope 46Ti x Abundance of Isotope 46Ti) + (Mass of Isotope 48Ti x Abundance of Isotope 48Ti) + (Mass of Isotope 50Ti x Abundance of Isotope 50Ti)
Average Atomic Mass = (45.95263 amu x 70.900%) + (47.94795 amu x 10.000%) + (49.94479 amu x 19.100%)
Average Atomic Mass ≈ 32.672 amu + 4.795 amu + 9.548 amu
Average Atomic Mass ≈ 47.015 amu
Therefore, the average atomic mass of titanium on that planet is approximately 47.015 amu.
1. To determine the number of moles of Al(CN)3 in 199 g of the compound, you need to use the formula:
moles = mass / molar mass
In this case, the molar mass of Al(CN)3 can be calculated by summing the molar masses of each element in the compound:
Al: 1 * atomic mass of Al
C: 1 * atomic mass of C
N: 3 * atomic mass of N
Look up the atomic masses of each element and compute the molar mass of Al(CN)3. Then, plug in the values:
moles = 199 g / molar mass of Al(CN)3
2. To determine the number of moles of P in 96.0 mol of P4O10, you need to use the coefficient of P in the balanced chemical formula of P4O10. In this case, there are 4 moles of P for every 1 mole of P4O10:
moles of P = 96.0 mol * 4 moles of P / 1 mole of P4O10
3. To determine the number of moles of CO2 produced in the combustion reaction of C5H12, you need to use the stoichiometry of the balanced equation. According to the equation:
1 mole of C5H12 produces 5 moles of CO2
Since you have 4.71 moles of C5H12, you can calculate the moles of CO2 produced:
moles of CO2 = 4.71 mol C5H12 * 5 moles of CO2 / 1 mole of C5H12
4. To determine the number of grams of O2 needed to burn 51.0 g of C3H8 completely, you need to use the molar ratio provided in the balanced equation. From the balanced equation:
1 mole of C3H8 reacts with 5 moles of O2
First, calculate the moles of C3H8 using its molar mass:
moles of C3H8 = 51.0 g / molar mass of C3H8
Then, use the stoichiometry to find the moles of O2 required:
moles of O2 = moles of C3H8 * 5 moles of O2 / 1 mole of C3H8
Finally, convert the moles of O2 to grams using its molar mass:
grams of O2 = moles of O2 * molar mass of O2
5. To determine the grams of NaI needed to produce 69.9 g of I2, you need to use the molar ratio provided in the balanced equation. From the balanced equation:
2 moles of NaI produce 1 mole of I2
First, calculate the moles of I2 using its molar mass:
moles of I2 = 69.9 g / molar mass of I2
Then, use the stoichiometry to find the moles of NaI required:
moles of NaI = moles of I2 * 2 moles of NaI / 1 mole of I2
Finally, convert the moles of NaI to grams using its molar mass:
grams of NaI = moles of NaI * molar mass of NaI
6. To calculate the average atomic mass of titanium using the given isotopes, you need to multiply the mass of each isotope by its abundance, then sum up the values:
Average atomic mass = (mass of isotope 1 * abundance of isotope 1) + (mass of isotope 2 * abundance of isotope 2) + ... + (mass of isotope n * abundance of isotope n)
In this case, the calculation would be:
Average atomic mass of titanium = (45.95263 amu * 0.709) + (47.94795 amu * 0.1) + (49.94479 amu * 0.191)