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4 years ago

For each of the following

Chemistry

2 answers:

baherus [9]4 years ago

6 0

Answer:

1. Skeleton: Ca + H₂O ⇒ H₂ + Ca(OH)₂

Balanced: Ca + 2H₂O ⇒ H₂ + Ca(OH)₂

Type: Single displacement / single replacement

2. Skeleton: Al + O₂ ⇒ Al₂O₃

Balanced: 4Al + 3O₂ ⇒ 2Al₂O₃

Type: Synthesis

3. Skeleton: H₂SO₄ + NaOH ⇒ Na₂SO₄ + H₂O

Balanced: H₂SO₄ + 2NaOH ⇒ Na₂SO₄ + 2H₂O

Type: Double displacement / double replacement

4. Skeleton: NaClO₃ ⇒ NaCl + O₂

Balanced: 2NaClO₃ ⇒ 2NaCl + 3O₂

Type: Decomposition

5. Skeleton: AgNO₃ + K₃PO₄ ⇒ Ag₃PO₄ + KNO₃

Balanced: 3AgNO₃ + K₃PO₄ ⇒ Ag₃PO₄ + 3KNO₃

Type: Double displacement / double replacement

6. Skeleton: Cu(OH)₂ + Al ⇒ Al(OH)₃ + Cu

Balanced: 3Cu(OH)₂ + 2Al ⇒ 2Al(OH)₃ + 3Cu

Type: Single replacement / single displacement

7. Skeleton: Mg + P₄ ⇒ Mg₃P₂

Balanced: 6Mg + P₄ ⇒ 2Mg₃P₂

Type: Synthesis

8. Skeleton: KNO₃ + PbI₂ ⇒ KI + Pb(NO₃)₂

Balanced: 2KNO₃ + PbI₂ ⇒ 2KI + Pb(NO₃)₂

Dimas [21]4 years ago

4 0

Answer:

1) Calcium metal and water react, giving hydrogen gas and calcium hydroxide

• Ca(s) + H2O Ca(OH)2 (aq) + H2(g)

• Ca(s) + 2H2O Ca(OH)2 (aq) + H2(g)

• Exothermic combination reaction

2) aluminum metal quickly reacts with the oxygen in the air to produce aluminum oxide

• Al + 02 = Al2O3

• 4 Al + 302 = 2Al2O3

• Combination reaction

3) Hydrogen sulphate (sulphutic acid) and sodium hydroxide react, producing sodium sulphate and water.

• NaOH + H2SO4 = Na2SO4 + 2H2O

• 2NaOH + H2SO4 = Na2SO4 + 2H2O

• Double Displacement reaction

4) sodium chloride and oxygen are produced by heating sodium chlorate.

• NaclO3 = NaCl + O

• 2NaclO3 = 2NaCl + 3 O

• Decomposition reaction

6) aluminum oxide and copper metal are the products of a reaction between copper (II) oxide and aluminum metal.

• Al + CuO = Al2O3 + Cu

• 2 Al + 3CuO = Al2O3 + 3 Cu

• Oxidation and Reduction

8) lead (II) nitrate and potassium iodide react producing lead (II) iodide a bright yellow precipitate and potassium nitrate which stays in solution.

• Pb(No3)2 +KI = PbI2 + KNO3

• Pb(No3)2 +2KI = PbI2 + 2KNO3

• Double displacement reaction

Explanation:

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Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$