Eric's father asked an engineer to survey the field behind their house. He wanted to plant some apple and pear trees there. Acco
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Mass of H₂ needed to react with O₂ : 1.092 g
<h3>Further explanation</h3>The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.
Reaction
O₂(g) + 2H₂(g) → 2H₂O(g)
mass of O₂ : 8.75 g
mol O₂(MW=32 g/mol) :
From the equation, mol ratio of O₂ : H₂ = 1 : 2, so mol H₂ :
Mass H₂ (MW=2 g/mol) :
Answer:
a. The limiting reactant is
b. 0.73 g of carbon dioxide are formed.
c. The grams of excess reactant that do not participate in the reaction are 0333 g.
Explanation:
a)
You know the following reaction:
⇒
First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:
- Na: 23
- H: 1
- C: 12
- O: 16
To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:
:23+1+12+16*3=84 g/mol
:1*3+12*6+1*5+16*7= 192 g/mol
:12+16*2= 44 g/mol
:1*2+16= 18 g/mol
: 23*3+12*6+1*5+16*7= 258 g/mol
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of react with 1 mole of
Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:
You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:
If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?
grams of sodium bicarbonate= 1.8375 g
But to perform the experiment you have only 1.4 g of sodium bicarbonate. So <u><em>the limiting reagent is sodium bicarbonate</em></u>.
b)
As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.
Now, by stoichiometry of the reaction, you know that 3 moles of react with 3 mole of
. Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:
You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?
<u><em>grams of carbon dioxide= 0.73 g</em></u>
<u><em>Then, 0.73 g of carbon dioxide are formed.</em></u>
c)
As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.
To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?
grams of citric acid= 1.067 g
But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:
grams of excess reactant=1.4 g - 1.067 g
grams of excess reactant=0.333 g
<em><u>So the grams of excess reactant that do not participate in the reaction are 0333 g.</u></em>