Question

1.) 3.4 moles Magnesium are mixed with 5.6 moles of Hydrochloric Acid. How many moles of Hydrogen gas are produced?

Answer 1

Answer:

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.