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Degger [83]
3 years ago
8

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

Chemistry
1 answer:
love history [14]3 years ago
3 0

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)

<u>Equation:</u>

Mg + <u>2</u>HCl → MgCl₂ + H₂

<u>Reactant to Product ratio</u>

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}}

<u>Determine actual yield of reactants</u>

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)

<u>Equation:</u>

C₃H₈  +  <u>5</u>O₂ → <u>3</u>CO₂ + <u>4</u>H₂O

<u>Reactant to Product ratio</u>

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}}

<u>Determine actual yield of reactants</u>

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.

<u>Equation:</u>

CH₄  + <u>2</u>O₂ → CO₂ + <u>2</u>H₂O

<u>Reactant to Product ratio</u>

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

<u>Actual yield:</u>

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)

<u>Equation:</u>

Pb(NO₃)₂ + <u>2</u>KI → PbI₂ + <u>2</u>KNO₃

<u>Reactant to Product ratio</u>

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

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

<u>Actual yield:</u>

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

The answer is 7.4 moles of KNO₃.

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3 0
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PLEASE HELP FAST IM ABOUT TO FAIL PLEASE HELP PLEASE HELP FAST HELP HELP THIS IS DESPRATE
MariettaO [177]

1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.

2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.

3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.

4. 135 ml of stock HBr will be required to dilute the solution.

5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution

6. The pH of the resulting solution is 13.89

Explanation:

The formula used in solving the problems is

number of moles= \frac{mass}{atomic mass of one mole}      1st equation

molarity = \frac{number of moles}{volume}            2nd equation

Dilution formula

M1V1 = M2V2          3rd equation

1. Data given

mass of Lithium nitrate = 21.6 grams

atomic mass of on emole lithium nitrate = 68.946 gram/mole

Molarity is given as 1.3 M

VOLUME=?

Calculate the number of moles using equation 1

n = \frac{21.6}{68.946}

  = 0.313 moles of lithium nitrate.

volume is calculated by applying equation 2.

volume = \frac{0.313}{1.3}

            = 0.240 litres of water will be used.

2. Data given:

mass of HCl = 215.1 gram

atomic mass of HCl = 36.46 gram/mole

volume = 2 litres

molarity = ?

using equation 1 number of moles calculated

number of moles = \frac{215.1}{36.46}

number of moles of HCl = 5.899 moles

molarity is calculated by using equation 2

M = \frac{5.899}{2}

   = 2.9 M is the molarity of the solution of 2 litre HCl.

3. data given:

molarity of H2SO4 = 18 M

Solution to be made 250 ml of 6 M

USING EQUATION 3

18 x V1= 250 x 6

V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.

4. data given:

M1= 10M, V1 =?, M2= 3 ,V2= 450 ml

applying the equation 3

10 x VI = 3x 450

V1 = 135 ml of stock HBr will be required.

5. Data given:

V1 = 50 ml

  M1= 12 M

  V2=?

  M2= 4

applying the equation 3

50 x 12 = 4 x v2

V2 = 150 ml.

6. data given:

HCl + NaOH ⇒ NaCl + H20

molarity of NaOH = 0.525 M

volume of NaOH = 25 ml

molarity of acid HCl= 75 ml

volume of HCl = 0.335 ml

pH=?

Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres

moles of NaOH = 0.0131

moles of HCl= 0.025 moles

The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.

difference of moles = 0.0119  moles ( NaOH moles is more)

Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution

molarity = \frac{0.0119}{0.1}

             = 0.11 M (pOH Concentration)

14 = pH + pOH  

  pH  = 14 - 0.11

     pH    = 13.89

3 0
3 years ago
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