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
Answer:
THE MOLARITY IS 2.22 MOL/DM3
Explanation:
The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water
Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol
Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;
37.5 g of Na2S = 217 g of water
( 37.5 * 1000 / 217 ) g = 1000 g of water
So, 172.81 g/dm3 of the solution
So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass
Molarity = 172.81 g/dm3 / 78 g/mol
Molarity = 2.22 mol/dm3
The molarity of the solution is 2.22 mol/dm3
A single-displacement reaction, also known as asingle-replacement reaction, is a type of chemicalreaction<span> where an element reacts with a compound and takes the place of another element in that compound. This type of </span>reaction<span> is typically pictured like this: Here, A replaces B in the compound BC.</span>
