Ask question

Ask question

All categories

Maksim231197 [3]

3 years ago

8

The first step in the reaction of Alka–Seltzer with stomach acid consists of one mole of sodium bicarbonate (NaHCO3) reacting wi

th one mole of hydrochloric acid (HCl) to produce one mole of carbonic acid (H2CO3), and one mole of sodium chloride (NaCl). Using this chemical stoichiometry, determine the number of moles of carbonic acid that can be produced from 5 mol NaHCO3 and 9 mol HCl.

1 answer:

Gnesinka [82]3 years ago

8 0

Answer: 5 moles of $H_2CO_3$ can be produced from 5 mol NaHCO3 and 9 mol HCl.

Explanation:

The balanced chemical reaction is:

$NaHCO_3+HCl\rightarrow H_2CO_3+NaCl$

According to stoichiometry :

1 mole of $NaHCO_3$ use 1 mole of $HCl$

Thus 5 moles of $NaHCO_3$ use= $\frac{1}{1}\times 5=5moles$ of $HCl$

Thus $NaHCO_3$ is the limiting reagent as it limits the formation of product and $HCl$ is the excess reagent.

As 1 mole of $NaHCO_3$ give = 1 mole of $H_2CO_3$

Thus 5 moles of $NaHCO_3$ give = $\frac{1}{1}\times 5=5moles$ of $H_2CO_3$

5 moles of $H_2CO_3$ can be produced from 5 mol $NaHCO_3$ and 9 mol HCl.

You might be interested in

A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

alekssr [168]

Answer:

208.7°C was the initial temperature of the limestone.

Explanation:

Heat lost by limestone will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of limestone = $m_1=62.6 g$

Specific heat capacity of limestone = $c_1=0.921 J/g^oC$

Initial temperature of the limestone = $T_1=?$

Final temperature = $T_2$ =T = 51.9°C

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.0 g$

Specific heat capacity of water= $c_2=4.186 J/g^oC$

Initial temperature of the water = $T_3=23.1^oC$

Final temperature of water = $T_2$ =T = 51.9°C

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(62.6 g\times 0.921 J/g^oC\times (51.9^oC-T_1))=75.0 g\times 4.186 J/g^oC\times (51.9^oC-23.1^oC)$

$T_1=208.7^oC$

208.7°C was the initial temperature of the limestone.

7 0

3 years ago

What is the molarity of 0.50g of Na dissolved in a 1.5 L solution?

lara [203]

M = mols of solute / L of solution

Convert .50 g Na to mols Na by dividing by 22.99 and then dividing that by 1.5, which gives you a 0.014 M solution.

5 0

3 years ago

When baking soda (sodium bicarbonate or sodium hydrogen carbonate, NaHCO3) is heated, it releases carbon dioxide gas, which is r

butalik [34]

Answer:

There is 78.25g NaHCO3 required

Explanation:

Step 1: Balance the equation

2 NaHCO3 → Na2CO3 + CO2 + H20

For 2 moles of NaHCO3 consumed, there is produced 1 mole of Na2CO3, 1 mole of CO2 and 1 mole of H2O.

Step 2: calculating moles of CO2

mass of Co2 = 20.5g

Molar mass of CO2 = 44.01 g/mole

moles of CO2 = 20.5 / 44.01 = 0.4658 moles

Step 3: Calculating moles of NaHCO3

Since we have for 1 mole CO2 produced, there is 2 moles of NaHCO3 consumed.

To calculate number of moles of NaHCO3, we have to multiply the number of moles of CO2, by 2.

⇒ 0.4658 x2 = 0.9316 moles

Step 4: Calculating the mass of NaHCO3

mass of NaHCO3 = moles of NaHCO3 x Molar mass of NaHCO3

mass = 0.9316 moles x 84g/ mole = 78.25g NaHCO3

There is 78.25g NaHCO3 required

4 0

3 years ago

What volume of stock solution and water must you add to prepare 36.25ml of a 1.25M solution

Serjik [45]

Given :

Number of moles , n = 36.25 mol .

Molarity , M = 1.25 M .

To Find :

The volume of water required .

Solution :

Moarity is given by :

$M=\dfrac{n}{V}$

So , $V=\dfrac{n}{M}$

Here , n is number of moles and M is molarity .

Putting all values in above equation , we get :

$V=\dfrac{36.25}{1.25}\\\\V=29\ L$

Therefore , volume of water required is 29 L .

5 0

3 years ago

Which compounds are the Bronsted- Lowry bases in the equilibrium?

Butoxors [25]

Its D for plato.
by other people who asked

4 0

3 years ago

Read 2 more answers