Interesting problem. Thanks for posting.
C2H2 + (3/2)02 ====> H2O + 2CO2
CH4 + 2O2 =====> 2H2O + CO2
The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16
The number of moles of C2H2 = x
The number of moles of CH4 = y
26x + 16y = 230.9 grams
For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18
x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water.
Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
From C2H2 we get 2*44*x = 88x grams of CO2
From CH4 we get 1*y*44 grams of CO2
88x + 44y for CO2
Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y = 972.7
Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7
I'm not going to go through the math unless you request me to do so.
x = 8.03 moles
y = 1.38 moles
The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.
1 mol of any particles has 6.02 * 10 ²³ particles.
If we look at 1 NH3 (1 mol NH3 or 1 molecule NH3), we can see that 1 molecule NH3 has 1 atom of N and 3 atoms of H; also 1 mole of NH3 has 1 mole of N atoms and 3 moles of H atoms.
So, 1 mol of NH3 has 1 mol of N atoms,
and 2.79 mol NH3 have 2.79 mol of N atoms.
2.79 mol of N atoms* 6.02 * 10 ²³ N atoms/ 1 mol N atoms = 1.68*10²⁴ N-atoms
Answer is 1.68*10²⁴ N-atoms.
Answer:
Element with 6s subshell
Explanation:
Reactivity of an element depends on the electronic configuration and position of element in the periodic table as reactivity increases as we go down the periodic table.
This is so because number of shell increases as move down the periodic table and the last electron is further away from the nucleus.
Element with 6s subshell is the largest among 3s and 4s subshell and has more number of shells so it will react more than 3s and 4s subshell.
Hence, the correct answer is "Element with 6s subshell".
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
