Chemistry, 01.01.2020 15:31 MoparorNocar061401

1.) 3.4 moles magnesium are mixed with 5.6 moles of hydrochloric acid. how many moles of hydrogen gas are produced?

2. ) 3.4 moles of c3h8 are burned in the presence of 12.3 moles of oxygen gas. how many moles of carbon dioxide are produced?

3.) 5.3 moles of water (h2o) are produced from the combustion of methane (ch4). how many moles of oxygen gas are required to produce the given amount of water?

4.) 3.7 moles of lead (ii) nitrate are mixed with 7.8 moles of potassium iodide. how many grams of potassium nitrate will be produced?

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Answer from: haley4023

1. 2.8 moles of H₂

2. 7.38 moles of CO₂

3. 5.3 moles of O₂

4. 7.4 moles of KNO₃

Explanation:

Here are the steps to doing this:

1. Write the chemical equation of each reaction.

2. Balance the equation.

3. Find out the ratio between reactant and product

4. Determine the actual yield of your reactants.

5. The amount of product produced is determined by how much product the limiting reactant produces.

Let's do this!

1. Given: 3.4 moles of Magnesium(Mg) and 5.6 moles of Hydrochloric acid (HCl)

Equation:

Mg + 2HCl → MgCl₂ + H₂

Reactant to Product ratio

1 mole of Mg produces 1 mole of H₂ $\dfrac{1moleof Mg}{1mole of H_{2}}$

2 moles of HCl produces 1 mole of H₂ $\dfrac{2molesofHCl}{1mole of H_{2}}$

Determine actual yield of reactants

$3.4moles of Mg\times\dfrac{1moleofH_2}{1moleofMg}=3.4molesofH_{2}\\\\5.6moles ofHCl\times\dfrac{1moleofH_2}{2moleofHCl}=2.8molesofH_{2}$

Since 5.6 moles of HCl can only produce 2.8 moles of H₂, before it is used up, then this means that that is all the product this reaction can produce.

2. Given: 3.4 moles of C₃H₈ and 12.3 moles of oxygen gas (O₂)

Equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Reactant to Product ratio

1 mole of C₃H₈ produces 3 moles of CO₂ $\dfrac{1moleofC_{3}H_{8}}{3molesofCO_{2}}$

5 moles of O₂ produces 3 moles of CO₂ $\dfrac{5molesofO_{2}}{3moleofCO_{2}}$

Determine actual yield of reactants

$3.4molesofC_{3}H_{8}\times\dfrac{3molesofCO_{2}}{1moleofC_{3}H_{8}}=10.2molesofH_{2}$

$12.3molesofO_{2}\times\dfrac{3molesofCO_{2}}{5molesofO_{2}}=7.38molesofCO_{2}$

The answer is then 7.38 moles of CO₂

**3. 5.3 moles of H₂O

This one is a little bit different. It is asking how much of a reactant is needed to produce the amount of product given. For this, just write a balanced equation for the reaction and get the ratio of reactant to product and solve for the actual yield. Since it is only asking for oxygen gas, you just need to do that one.

Equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Reactant to Product ratio

$\dfrac{2molesofO_{2}}{2molesofH_{2}O}=\[tex]7.88molesofKI\times\dfrac{1moleofKNO_{3}}{1moleofKI}=7.88molesofKNO_{3}$

Actual yield:

$5.3molesofH_{2}O\times\dfrac{1moleofO_{2}}{1moleofH_{2}O}=5.3molesofO_{2}$

The answer is 5.3 moles of O₂.

4. 3.7 moles of Lead (II) Nitrate (Pb(NO₃)₂) and 7.8 moles of Potassium Iodide (KI)

Equation:

Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃

Reactant to Product ratio

$\dfrac{2molesofKI}{2molesofKNO_{3}}=\dfrac{1moleofKI}{1moleofKNO_{3}}$

$\dfrac{1molesofPb(NO_{3})_{2}}{2molesofKNO_{3}}$

Actual yield:

$3.7molesofPb(NO_{3})_{2}\times\dfrac{2molesofKNO_{3}}{1moleofPb(NO_{3})_{2}}=7.4molesofKNO_{3}$

The answer is 7.4 moles of KNO₃.

Answer from: Quest

a rain shadow is a dry region of land on the side that is protected against the prevailing winds of a mountain range. the prevailing winds are the winds that occur most of the time in a particular place of the earth

prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of rain or snow. this dry air produces a rain shadow.

Answer from: Quest

i think it is true since you can test the answer and can use observations to complete it.

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