Answer:
NO will be the limiting reagent.
Explanation:
The balanced equation is:
2 NO + 2 CO → N₂ + 2 CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- NO: 2 moles
- CO: 2 moles
- N₂: 1 mole
- CO₂: 2 moles
Being the molar mass of each compound:
- NO: 30 g/mole
- CO: 28 g/mole
- N₂: 28 g/mole
- CO₂: 44 g/mole
Then by stoichiometry the following quantities of mass participate in each reaction:
- NO: 2 moles* 30 g/mole= 60 g
- CO: 2 moles* 28 g/mole= 56 g
- N₂: 1 mole* 28 g/mole= 28 g
- CO₂: 2 moles* 44 g/mole= 88 g
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, you can use a simple rule of three as follows: If 56 grams of CO react with 60 grams of NO, 3 grams of CO react with how much mass of NO?
mass of NO= 3.21 grams
But 3.21 grams of NO are not available, 3 grams are available. Since you have less moles than you need to react with 3 grams of CO, <u><em>NO will be the limiting reagent.
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Answer:
The answer is
<h2>0.0056 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
volume = 3.4 mL
1 mg = 0.001 g
18.99 mg = 0.01899 g
So we have
We have the final answer as
<h3>0.0056 g/mL</h3>
Hope this helps you
Answer:
True, if you slice an apple in half, you have 1/2 of an apple.
<h3>
Answer:</h3>
0.89 J/g°C
<h3>
Explanation:</h3>
Concept tested: Quantity of heat
We are given;
- Mass of the aluminium sample is 120 g
- Quantity of heat absorbed by aluminium sample is 9612 g
- Change in temperature, ΔT = 115°C - 25°C
= 90°C
We are required to calculate the specific heat capacity;
- We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature.
That is;
Q = m × c × ΔT
- Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium.
Specific heat capacity, c = Q ÷ mΔT
= 9612 J ÷ (120 g × 90°C)
= 0.89 J/g°C
Therefore, the specific heat capacity of Aluminium is 0.89 J/g°C