Calculate the MASS if the density is 2.34 g/mL and the volume is 12.50 mL.

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:

a)

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 the limiting reagent is sodium bicarbonate.

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 $NaHCO_{3}$ 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?

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

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

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?

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

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

So the grams of excess reactant that do not participate in the reaction are 0333 g.

3 0

3 years ago

What is the appearance of the signal corresponding to the CH₂ protons in the 1H NMR spectrum of ethyl methyl ether, CH₃CH₂OCH₃?

Tems11 [23]

Answer: option C. a quartet

Explanation:

7 0

3 years ago

A gas mixture is made up of kr (21.7 g), o2 (7.18 g), and co2 (14.8 g). the mixture has a volume of 23.1 l at 59 °c. calculate t

Scilla [17]

Answer:- partial pressure of Kr = 0.306 atm, partial pressure of oxygen = 0.264 atm and partial pressure of carbon dioxide = 0.396 atm

Total pressure is 0.966 atm

Solution:- moles of Kr = 21.7 g x (1mol/83.8g) = 0.259 mol

moles of oxygen = 7.18 g x (1mol/32g) = 0.224 mol

moles of carbon dioxide = 14.8 g x (1mol/44g) = 0.336 mol

Volume of container = 23.1 L and the temperature is 59 + 273 = 332 K

From ideal gas law equation, P = nRT/V

partial pressure of Kr = (0.259 x 0.0821 x 332).23.1 = 0.306 atm

partial pressure of oxygen = (0.224 x 0.0821 x 332)/23.1 = 0.264 atm

partial pressure of carbon dioxide = (0.336 x 0.0821 x 332)/23.1 = 0.396 atm

Total pressure of the gas mixture = 0.306 atm + 0.264 atm + 0.396 atm = 0.966 atm

4 0

3 years ago

Predict the products butane (C4H10) + oxygen —>

zloy xaker [14]

C4H10+ 02= C02+ H20

explanation

a combusting is made of carbon dioxide and water.

C=4 C=1 x4 =4

H=10 H=2x5=10

O=2 x13/2=13 O = 8+5=13

6 0

3 years ago

B. Radiation

Vika [28.1K]

First one, occurs fast !

4 0

3 years ago

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