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3 years ago

How many moles of water are made from complete reaction of 1.4 moles of hydrogen gas? Given the reaction: 2H2 + O2 → 2H2O

Chemistry

1 answer:

Ber [7]3 years ago

8 0

Answer:

1.4moles

Explanation:

From the balanced equation below;

2H2 + O2 → 2H2O

2moles of hydrogen reacts with 1mole of oxygen to produce 2moles of water.

since we are only interested in comparing hydrogen and water, we can say 2moles of hydrogen will produce 2moles of water

Hence by comparism of moles;

1.4moles of hydrogen will produce 1.4moles of water

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Consider two flasks at 25 degree Celsius, one contains an ideal gas and the other contains the real gas SO3. Which statement reg

Strike441 [17]

Answer: Option (A) is the correct answer.

Explanation:

In real gases, there exists force of attraction between the molecules at low temperature and high pressure. This is because at low temperature there occurs a decrease in kinetic energy of gas molecules and high pressure causes the molecules to come closer to each other.

As a result, forces of attraction increases as molecules come closer to each other and therefore, gases deviate from an ideal gas behavior.

And, at low pressure and high temperature there exists no force of attraction or repulsion between the molecules of a gas because they have high kinetic energy. Hence, gases behave ideally at these conditions.

Thus, we can conclude that the statement as the temperature approaches 0 K, the volume of the ideal gas will be larger than the volume of $SO_{3}$ because ideal gases lack inter-molecular forces, is true.

3 0

2 years ago

Type the correct answer in the box. Express your answer to three significant figures.

Shkiper50 [21]

Answer: Theoretical mass of sodium sulphate ( $Na_2SO_4$ ) is 514.118 grams.

Explanation: For a given reaction,

$2NaOH(aq.)+H_2SO_4(aq.)\rightarrow Na_2SO_4(s)+2H_2O(l)$

As NaOH is used in excess, therefore it is an excess reagent and $H_2SO_4$ is a limiting reagent as the quantity of the product will depend on it.

We are given 355 grams of $H_2SO_4$ .

Molar mass of $H_2SO_4$ = 98.079 g/mol

Molar mass of $Na_2SO_4$ = 142.04 g/mol

1 mole of $H_2SO_4$ is producing 1 mole of $Na_2SO_4$ , so

98.079 g/mol of $H_2SO_4$ will produce 142.04 g/mol of $Na_2SO_4$

355 grams of $H_2SO_4$ will produce = $142.04g/mol \times \frac{355g}{98.079g/mol}$ of $Na_2SO_4$

Mass of $Na_2SO_4$ = 514.118 grams

3 0

3 years ago

Read 2 more answers

Tarnished silver can be cleaned with toothpaste. Give reason

vfiekz [6]

Toothpaste is an abrasive that contains many things, but mainly phosphates.

6 0

2 years ago

For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will r

s344n2d4d5 [400]

Answer:

a) 26.67 moles HNO3

b) 0.33 moles NO

c) 0.40 moles NO is produced

d).157 moles Cu

e) 0.105 moles NO

f) 26.4 grams HNO3

g) Cu is in excess

h) 2.41 grams Cu remain

i) 2.37 grams NO

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = 26.67 moles HNO3

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = 0.33 moles NO

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = 0.40 moles NO is produced

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have 0.157 moles Cu

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = 0.105 moles NO

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = 26.4 grams HNO3

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). Cu is in excess. There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol = 2.37 grams NO

3 0

3 years ago

How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac

Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

• using the molar mass of Ca(OH)2 (74g/mol).

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

• using the molar mass of HNO3 (63.02g/mol).

$4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3$

So, 7.32g of HNO3 are required.

4 0

1 year ago