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kondor19780726 [428]
3 years ago
13

When copper (II) chloride reacts with sodium nitrate, copper (II) nitrate and sodium chloride are formed.

Chemistry
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

How many grams of copper (II) nitrate is formed

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The temperature of evaporation is much higher for water than for alcohol. Without knowing more about the chemistry of alcohol, w
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Hydrogen bonding is a kind of intermolecular interaction that occurs when hydrogen is bonded to a highly electronegative atom.

Both water and alcohols exhibit hydrogen bonding. However, alcohols exhibit fewer hydrogen bonds than water.

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1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in
FrozenT [24]

Answer:

1)  <u>10.0 moles of NO</u>

<u>2) 25 moles of NaCl</u>

3) <u>1200 moles of CO2</u>

<u>4) 1.03 moles of MgO</u>

<u>5) 0.72 moles H2</u>

<u>6) 1041.15 grams BaCl2</u>

<u>7) </u>9.55 grams MgO

8) <u>45.5 grams Au</u>

<u>9 )14.93 grams AlCl3</u>

Explanation:

1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in the following composition reaction?

N2 + O2 → 2NO

For 1 mol N2 we need 1 mol O2 to produce 2 moles of NO

For 5.0 moles of N2 we need 5.0 moles of O2 to produce <u>10.0 moles of NO</u>

2. The neutralization of an acid with a base is a double replacement reaction in which a salt and water are formed. If you start with 25 moles of HCl and neutralize it with  NaOH how many moles of NaCl will be formed?

HCl + NaOH → NaCl + H2O

For 1 mol HCl we need 1 mol NaOH to produce 1 mol of NaCl and 1 mol H2O

For 25 moles of HCl we need 25 moles of NaOH to produce <u>25 moles of NaCl</u> and 25 moles of H2O

3. A car burns gasoline (octane – C8H18) with oxygen. If you drive to Salt Lake and  burn 150 moles of octane how many moles of carbon dioxide are you producing?

2C8H18 + 25O2 → 16CO2 + 18H2O

For 2 moles of octane we need 25 moles of O2 to produce 16 moles of CO2 and 18 moles of H2O

For 150 moles of octane we need 25*75 = 1875 moles of O2

To produce 16*75 = <u>1200 moles of CO2</u> and 18*75= 1350 moles

4. If 25 gram of magnesium combines with oxygen in a composition reaction, how  many moles of magnesium oxide will be formed?

2Mg + O2 → 2MgO

Moles of Mg = 25.0 g/24.3 g/mol = 1.03 moles

For 2 moles Mg we need 1 mol O2 to produce 2 moles MgO

For 1.03 moles Mg we'll have <u>1.03 moles of MgO</u>

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5. . Lithium reacts with water in a single replacement reaction. How many moles of  hydrogen gas a produced by 10 grams of lithium?

2Li + 2H2O → 2LiOH + H2

Moles Li = 10.0 grams/ 6.94 g/mol = 1.44 moles

For 2 moles Li we need 2 mole H2O to produce 2 moles LiOH and 1 mol H2

For 1.44 moles Li we need 1.44 moles H2O to produce 1.44 moles H2O and <u>0.72 moles H2</u>

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6. Barium chloride reacts with sodium sulfate in a double replacement reaction. How  many grams of barium chloride are required to react with 5 moles of sodium sulfate?

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

For 1 mol of BaCl2 we need 1 mol of Na2SO4 to produce 1 mol of BaSO4 and 2 moles NaCl

For 5 moles Na2SO4 we need 5 moles BaCl2

mass BaCl2 = 5 moles * 208.23 g/mol = <u>1041.15 grams BaCl2</u>

7. Magnesium carbonate when heated decomposes to form magnesium oxide and carbon dioxide. How many grams of magnesium oxide will be formed if 20 grams of  magnesium carbonate are heated?

MgCO3 → MgO + CO2

Moles MgCO3 = 20.0 grams / 84.31 g/mol

Moles MgCO3 = 0.237 moles

For 1 mol MgCO3 we'll have 1 mol MgO and 1 mol CO2

For 0.237 moles MgCO3 we'll have 0.237 moles MgO and 0.237 moles CO2

Mass MgO = 0.237 moles * 40.30 g/mol = 9.55 grams MgO

8. If 70 grams of gold III chloride decomposes into its elements, how many grams of  gold will be produced?

2AuCl3 → 2Au + 3Cl2

Moles AuCl3 = 70 grams / 303.33 g/mol = 0.231 moles

For 2 moles AuCl3 we'll have 2 moles gold and 3 moles Cl2

For 0.231 moles AuCl3 we'll have 0.231 moles gold

Mass of gold =  0.231 moles * 196.97 g/mol = <u>45.5 grams Au</u>

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9. Chlorine is more reactive element than bromine, thus chlorine will replace bromine in compound through a single replacement reaction. If 30 grams of aluminum bromide react with chlorine in this fashion how many grams of aluminum chloride will be formed?

2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2

Moles AlBr3 = 30 g /266.69 g/mol = 0.112 moles

For 2 moles AlBr2 we need 3 moles Cl2 to produce 2 moles AlCl3 and 3 moles Br2

For 0.112 moles AlBr3 we need 3/2 * 0.112 = 0.168 moles of Cl2

To produce 0.112 moles of AlCl3 and 0.168 moles of Br2

Mass AlCl3 = 0.112 moles * 133.34 g/mol = <u>14.93 grams AlCl3</u>

8 0
3 years ago
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