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MissTica
3 years ago
7

How many grams of water are produced if we react 3 moles of hydrogen with 3 moles of oxygen?

Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
4 0

Answer:

About 60 grams

Explanation:

The balanced equation of the reaction is given as;

2H2(g) + O2(g) → 2H2O(l)

From the equation;

2 mol of H2 reacts with 1 mol of O2

3 mol of H2 would require 3/2 mol of O2 (considering the 2:1 ratio)

The limiting reactant is H2 as it would be used up before H2. It determines the amount of product that would be formed.

2 mol of H2 produces 2 mol of H2O

3 mol of H2 would produce 3 mol of H2O (Considering the 1 : 1 ratio)

Converting moles to mass;

Mass = Molar mass * Number of moles

Mass = 18 * 3 = 54 g

The correct option is; About 60 grams

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Provide the IUPAC names for
ollegr [7]

<u>Answer:</u>

<u>For a:</u> The IUPAC name of the compound is N-ethylethaneamide.

<u>For b:</u> The IUPAC name of the compound is N,N-diethylmethaneamide.

<u>For c:</u> The IUPAC name of the compound is ethyl pentanoate

<u>Explanation:</u>

To name a compound, first look for the longest possible carbon chain.

  • <u>For a:</u>

Amide group is a type of functional group where an amine group is attached to a carbonyl group. The general formula of amide is R-CO-NH_2, where R is an alkyl or aryl group.

In part (a), the alkyl group has 2 carbon atoms and thus, the prefix used is 'eth-'

Also, an ethyl substituent is directly attached to N-atom. It is an alkane structured hydrocarbon thus, the suffix used will be '-ane'

Hence, the IUPAC name of the compound is N-ethylethaneamide.

  • <u>For b:</u>

Amide group is a type of functional group where an amine group is attached to a carbonyl group. The general formula of amide is R-CO-NH_2, where R is an alkyl or aryl group.

In part (b), the alkyl group has 1 carbon atoms and thus, the prefix used is 'meth-'

Also, two ethyl substituents are directly attached to N-atom. It is an alkane structured hydrocarbon thus, the suffix used will be '-ane'

Hence, the IUPAC name of the compound is N,N-diethylmethaneamide.

  • <u>For c:</u>

Esters are a kind of organic molecules having functional groups, R-COO-R' where R and R' are the alkyl or aryl groups. They are formed by the combination of alcohol and carboxylic acid.

These functional group compounds are named in two words which is alkyl alkanoates, where alkyl refers to the alcoholic part and alkanoate refers to the carboxylic acid part of the molecule. The numbering of the parent chain in esters is done from the carboxylic carbon. The alkyl part is not given any numbers.

In part (c), there are 5 carbon atoms present in a straight chain and thus, the prefix used is 'pent-'

Also, an ethyl group forms the alcoholic part.

Hence, the IUPAC name of the compound is ethyl pentanoate

6 0
3 years ago
Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a
Marat540 [252]

Answer : The partial pressure of helium is, 1.815\times 10^4KPa

Solution : Given,

Molar mass of O_2 = 32 g/mole

Molar mass of helium = 4 g/mole

Molar mass of N_2 = 28 g/mole

Total pressure of gas = 2.07\times 10^4KPa

As we are given gases in percent, that means 10 g of oxygen gas, 50 g of helium gas and 40 g of nitrogen gas present in 100 g of mixture.

First we have to calculate the moles of oxygen, helium and nitrogen gas.

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{10g}{32g/mole}=0.3125moles

\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{50g}{4g/mole}=12.5moles

\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{40g}{28g/mole}=1.428moles

Now we have to calculate the total number of moles of gas mixture.

\text{Total number of moles of gas}=\text{Moles of oxygen gas}+\text{Mole of helium gas}+\text{Moles of nitrogen gas}

\text{Total number of moles of gas}=0.3125+12.5+1.428=14.24moles

Now we have to calculate the moles fraction of helium gas.

\text{Mole fraction of He gas}=\frac{\text{Moles of He gas}}{\text{Total number of moles of gas}}=\frac{12.5}{14.25}=0.877

Now we have to calculate the partial pressure of helium.

p_{He}=X_{He}\times P_T

where,

p_{He} = partial pressure of helium

P_T = total pressure

X_{He} = mole fraction of helium

Now put all the given values in this formula, we get

p_{He}=(0.877)\times (2.07\times 10^4KPa)=1.815\times 10^4KPa

Therefore, the partial pressure of helium is, 1.815\times 10^4KPa

8 0
3 years ago
How does chemical change affect the composition of matter
Sergio039 [100]
Beacuse it gets oxagin and changes ur wlecome

8 0
3 years ago
PLEASE HELP FAST !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
MakcuM [25]
The answer is D: 400N
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5 0
3 years ago
0.352 g sample of a diprotic acid is dissolved in water and titrated with 0.150 M NaOH.0.150 M NaOH. What is the molar mass of t
Aliun [14]
<h3>Answer:</h3>

128.94 g/mol

<h3>Explanation:</h3>

Given;

Mass of diprotic acid, H₂X =0.352 g

Molarity of NaOH = 0.150 M

Volume of the NaOH = 36.4 mL

We are required to calculate the molar mass of the acid.

Note: A diprotic acid is an acid that contains 2 replaceable hydrogen atoms

<h3>Step 1: Write the balanced equation fro the reaction;</h3>

The balanced equation for the reaction between the diprotic acid and NaOH will be;

H₂X(aq) + 2NaOH(aq) → Na₂X(aq) + 2H₂O(l)

<h3>Step 2: Determine the number of moles of NaOH used </h3>

Given the molarity and volume of NaOH we can calculate the number of moles;

Moles of NaOH = Molarity × Volume

                          = 0.150 M × 0.0364 L

                         = 0.00546 moles

<h3>Step 3: Use the mole ratio to determine moles of the acid </h3>

From the equation;

1 mole of the acid reacts with 2 moles of NaOH

Therefore; H₂x : NaOH = 1 : 2

Moles of H₂x = 0.00546 moles ÷ 2

                      = 0.00273 moles

<h3>Step 4: Determine the molar mass of the acid.</h3>

Molar mass is the mass equivalent to 1 mole of a compound or element

From our calculations;

0.00273 moles = 0.352 g of the acid;

Therefore, mass in 1 mole ;

= 0.352 g ÷ 0.00273 moles

= 128.94 g/mol

Thus, the molar mass of the diprotic, H₂X is 128.94 g/mol

7 0
3 years ago
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