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
You can have a solution of hydrogen peroxide that might say 10% that means that 10% per mass of the hydrogen peroxide solution is the hydrogen peroxide the rest is water.
concentration is the amount of mass in the solution eg 5gdm-3
hope that helps
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Answer:
The answer to your question is: letter D.
Explanation:
Noble gases are located in group VIIIA of the periodic table, this means that they have 8 eight electrons in their outermost shell.
Due to this characteristic, they are stable and do not react with other elements.
a. 1s22s22p4 The outermost shell of this electron configuration has 6 electrons, then this element has 6 electrons not 8. This configuration is of an element of the group VIA.
b. [Ne]2s22p2 The outermost shell of this element has 4 electrons, so this is not the configuration of a noble gas.
c. [Ar] 3s1 This element only has one electron in its outermost shell, so this is the electron configuration of an alkaline metal.
d. 1s22s22p6 This element has 8 electrons in its outermost shell, so this is the electron configuration of a noble gas.
Hello!
To solve this problem, we will use the
Boyle's Law, which describes how pressure changes when volume changes and vice-versa. The equation for this law is the following one, and we'll clear for V2:
So, the final volume after increasing the pressure would be
2,7 L. That means that volume decreases when the pressure increases
Have a nice day!
