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3 years ago

What mass of water can be obtained from 4.0 g of H2 and 16 g of O2?2 H2 + O2 ---> 2 H2O18 g36 g54 g9 g

Chemistry

2 answers:

SashulF [63]3 years ago

7 0

Answer:

18 g is the mass produced by 4 g of H₂ and 16 g of O₂

Explanation:

The reaction is:

2H₂ + O₂ → 2H₂O

So, let's find out the limiting reactant as we have both data from the reactants.

Mass / Molar mass = moles

4 g/ 2g/m = 2 moles H₂

16g / 32 g/m = 0.5 moles O₂

2 moles of hydrogen react with 1 mol of oxygen, but I have 0.5, so the O₂ is the limiting.

1 mol of O₂ produces 2 mol of water.

0.5 mol of O₂ produce (0.5 .2)/1 = 1 mol of water.

1 mol of water weighs 18 grams.

madreJ [45]3 years ago

7 0

Answer:

18 grams of $H_2O$

Explanation:

The balanced equation of the reaction is:

$H_2+\frac{1}{2}O_2 -->H_2O$

From the balanced equation we can say 1 mole of H2 reacts with 0.5 moles of O2 to give one mole of H2O.

Number of moles of H2 = $\frac{Given\ mass}{Molar\ mass}=\frac{4}{2}=2\ moles$

Number of moles of O2 = $\frac{Given\ mass}{Molar\ mass}=\frac{16}{32}=0.5\ moles$

We have 2 moles H2 and 0.5 moles of O2.

Not all H2 reacts because the amount of O2 is limited.

Since only 0.5 moles of O2 is available only 1 mole of H2 reacts according to the balanced equation.

Hence 1 mole of H2O is formed which is 18 grams.

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A solution is 40 cetic acid by mass. the density of this solution is 1.049 g/ml. Calculate the mass of pure acetic acid in 220 m

taurus [48]

The mass of pure acetic acid in 220 ml of the given solution at 20°C is 92.311 g

<h3>What is Acetic acid?</h3>

Acetic acid is a type of carboxylic acid and also known as ethanoic acid

Its formula is CH₃COOH.

It is an organic compound and is a colorless liquid

It is mostly used in the production of vinegar

40 % acetic acid by mass means,

40 g of acetic acid is dissolved in 100 g of solution.

The density of solution at 20°C,

$\rho = 1.049 g/ml$

We know,

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

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

The volume of the solution, $V = \frac{100}{1.049}$ = 95.33 ml

95.33 ml of solution contains 40 g of pure Acetic acid

220 ml of solution contains $\frac{40 \times 220}{95.33}$ = 92.311 g of pure Acetic acid

Thus, the mass of pure Acetic acid in 220 ml of solution at 20°C is 92.311 g

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2 years ago

A compound has an empirical formula of CH2 and a molar mass of 56 g. What is its molecular formula?

kipiarov [429]

Answer:

C4H8

Explanation:

First find the molar mass of CH2;

2(1.01) + 1(12.01) = 14.03g

Now divide the molar mass of the compound by the molar mass of CH2;

56g/14.03g = 3.9914 Round to nearest whole number = 4

Multiply CH2 by 4 to get the molecular formula;

CH2* 4 = C4H8

4 0

3 years ago

Draw an alkyl bromide with proper stereochemistry that can be used to synthesize the given alkene as the exclusive product via a

enot [183]

Answer:

See explanation below

Explanation:

You forgot to put the picture to do so. In this case, I manage to find one, and I hope is the one you are looking for. If not, then post it again and I'll gladly help you out again.

According to the picture with the answer, we have a cyclohexane with 4 methyl groups there. Two of them are facing towards the molecule with a darker bond. This means that the alkyl bromide, should have a bromine in one of the bonds, and in order to produce an E2 reaction, this bromine should be facing in the opposite direction of the methyl groups which are facing towards. This is because an E2 reaction occurs with the less steric hindrance in the molecule. If the bromine is in the same direction as the methyl group, it will cause a lot more of work to do a reaction, and therefore, an E2 reaction. I will promote instead a E1 or a sustitution product.

Therefore the alkyl bromide should be like the one in the picture 2.

3 0

3 years ago

How many grams are in 2.3 moles of N?

jok3333 [9.3K]

The amount of grams that are in 2.3 moles of N = 32.223 or 32/100
Because there are 14.01 grams per mile of nitrogen atoms.
So…
14.01 x 2.3= 32.223
Hope this helps :)

6 0

2 years ago

Read 2 more answers

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago