All categories

1 year ago

(b) What amount (mol) of H₂ can be produced from the given mass of H₂O?

Chemistry

1 answer:

Luden [163]1 year ago

8 0

8 moles of H 2O are produced.

First, we need to figure out the chemical equation for producing water with oxygen which is H 2 + O2 = H 2O. Then, we need to balance the equation, resulting in 2H 2 + O2 = 2H 2O.

<h3>How many moles of H2 are required to make one mole of NH3?</h3>

Calculate 0.88074 mol H2's mass. If N2 is too much, 1.776 g H2 is needed to create 10.00 g of NH3. To create 8.2 moles of ammonia, 2 moles of NH3 are created when 1 mole of N2 and 3 moles of H2 mix. 4.1 moles of N2 Fast are consequently needed to make 8.2 moles of NH3.

<h3>How many moles of h2 are needed to produce a solution?</h3>

An O-H bond has a bond energy of 1 09 Kcal. 3.6. A 38.0mL 0.026M HCl solution and a 0.032M NaOH solution react. Thus, 10 moles of NH 3 are obtained by dividing 15 moles of H2 by the 1.5 moles of H2 required for the product. and 9.3 x 10-3 moles of bromobutane (1.27/137 =.00927moles).

Learn more about H2O:

brainly.com/question/2193704

#SPJ4

You might be interested in

What type of change forms a new<br> substance?

marusya05 [52]

Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions. The “ingredients” of a reaction are called reactants, and the end results are called products.

6 0

2 years ago

If you start with 4.3 moles of NH3, how many moles of NANH2 can be produced

mash [69]

Explanation:

Mole ratio of NH3 to NaNH2 = 2 : 2.

Moles of NaNH2 produced

= 4.3mol * (2/2) = 4.3mol.

4 0

2 years ago

The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

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:

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

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:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 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 $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

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?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

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>.

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 $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

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?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

<u><em>grams of carbon dioxide= 0.73 g</em></u>

<u><em>Then, 0.73 g of carbon dioxide are formed.</em></u>

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?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

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>

3 0

3 years ago

you know the mass of an object and the force applied to the bject to make it move . Which of Newtons laws of motion will help yo

MAVERICK [17]

Answer:

second law of acceleration

Explanation:

hope this helps :)

8 0