CH4 is the ___________________ formula for methane.

Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H

allochka39001 [22]

Answer : The theoretical yield of water formed from the reaction is 0.54 grams.

Solution : Given,

Mass of HCl = 1.1 g

Mass of NaOH = 2.1 g

Molar mass of HCl = 36.5 g/mole

Molar mass of NaOH = 40 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of HCl and NaOH.

$\text{ Moles of }HCl=\frac{\text{ Mass of }HCl}{\text{ Molar mass of }HCl}=\frac{1.1g}{36.5g/mole}=0.030moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{2.1g}{40g/mole}=0.525moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NaOH

So, 0.030 mole of HCl react with 0.030 mole of NaOH

From this we conclude that, $NaOH$ is an excess reagent because the given moles are greater than the required moles and $HCl$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 1 mole of $HCl$ react to give 1 mole of $H_2O$

So, 0.030 moles of $HCl$ react to give 0.030 moles of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

$\text{ Mass of }H_2O=(0.030moles)\times (18g/mole)=0.54g$

Therefore, the theoretical yield of water formed from the reaction is 0.54 grams.

6 0

3 years ago

Iron has a density of 7.86 g/cm3. Calculate the volume (in dL) of a piece of iron having a mass of 4.07 kg . Note that the densi

elena-s [515]

Answer:

The volume of the piece of iron is 5.18dL.

Explanation:

The density (ρ) is equal to the mass (m) divided the volume (V).

$\rho = \frac{m}{V}$

If we rearrange it, we have:

$V=\frac{m}{\rho }$

To express the volume in dL we will need the following relations:

1 dL = 0.1 L
1 kg = 10³ g
1 cm³ = 1 mL
1mL = 10⁻³L

Then,

$\rho = \frac{7.86g}{cm^{3} } .\frac{1cm^{3} }{1mL} .\frac{1mL}{10^{-3}L } .\frac{1kg}{10^{3}g } =7.86kg/L$

Finally,

$V=\frac{m}{\rho }=\frac{4.07kg}{7.86 kg/L} =0.518L.\frac{1dL}{0.1L} =5.18dL$

5 0

4 years ago

Read 2 more answers

Ocean waves are limited in

IRINA_888 [86]

Answer:

are there answer choices with the question?

Explanation:

sorry im a little confused

5 0

4 years ago

4. Is the ketone oxidized or reduced during this reaction and how can you tell?

Lady_Fox [76]

The answer will be a

3 0

3 years ago

Carbon Monoxide is a toxic gas that is colorless and tasteless. How many moles of CO are in 1 L of the gas at STP?

Debora [2.8K]

Answer:

0.045 moles

Explanation:

Given data:

Moles of CO = ?

Volume of gas = 1 L

Temperature and pressure = standard

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

n = PV/RT

n = 1 atm× 1L / 0.0821 atm.L/mol.K × 273 K

n = 1 atm.L / 22.41 atm.L/mol

n = 0.045 mol

7 0

3 years ago