An incandescent bulb becomes hotter than a fluorescent bulb when turned on because in a regular incandescent bulb, there is tungsten wire where electricity is converts into heat. A regular incandescent light bulb requires 4 times more energy than a fluorescent bulb in order to produce the same amount of light. The conversion is such that for a 75-watt bulb, temperature get raised to approximately 2000 K. For such a high temperature, the radiating energy from the wire have some visible light. In such bulbs, 90% of the electricity get consumed in producing heat and only 10% produces light thus, they are not much efficient source of light.
On the other hand, fluorescent bulbs produce light with less amount of heat. In them, 40% of electricity is consumed in producing light and 60% in heat which is very less as compared to heat produced by a incandescent bulb. This is because when it get turned on, mercury atoms inside the bulb collides with electrons and produce UV light which is then converted into visible light using thin layer of phosphor power present inside the bulb. This produces low amount of heat thus, the bulb stays cooler, the bigger size of bulb also helps in dispersing heat.
Therefore, a fluorescent light bulb is not as hot as an incandescent light bulb.
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
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