Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
First, we calculate the mass of the sample:
mass = density x volume
mass = 8.48 x 112.5
mass = 954 grams
Now, we will calculate the mass of each component using its percentage mass, then divide it by its atomic mass to find the moles and finally multiply the number of moles by the number of particles in a mole, that is, 6.02 x 10²³.
Zinc mass = 0.37 x 954
Zinc mass = 352.98 g
Zinc moles = 352.98 / 65
Zinc moles = 5.43
Zinc atoms = 5.43 x 6.02 x 10²³
Zinc atoms = 3.27 x 10²⁴
Copper mass = 0.63 x 954
Copper mass = 601.02 g
Copper moles = 601.02 / 64
Copper moles = 9.39
Copper atoms = 9.39 x 6.02 x 10²³
Copper atoms = 5.56 x 10²⁴
Answer:
A cation is formed when a metal ion loses a valence electron while an anion is formed when a non-metal gains a valence electron. They both achieve a more stable electronic configuration through this exchange.
The energy of the carbide released is 7262.5MJ.
<h3>What is the energy?</h3>
We know that the reaction between calcium oxide and carbon occurs in accordance with the reaction; 
. The reaction is seen to produce 464.8kJ of energy per mole of carbide produced.
Number of moles of 
produced = 1000 * 10^3 g/64 g/mol
= 15625 moles of calcium carbide
If 1 mole of transfers 464.8 * 10^3 J
15625 moles of calcium carbide transfers 15625 moles * 464.8 * 10^3 J/ 1 mol
= 7262.5MJ
Learn more about reaction enthalpy:brainly.com/question/1657608
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I don’t know what exactly is but see at the picture.
