Question

Use stoichiometry to calculate the following questions 1) Plants utilize light energy in the photosynthesis process to synthesize glucose, C6H12O6, from CO2 and H2O by way of the reaction 6CO2 +6H2O- C¢H12O6 +602. How many grams of CO2 are consumed in the production of 90 g of glucose? 2) For the reaction of 100.0 g of NAOH with Cl2, 2NAOH +Cl2 ->NaClO + NaCI+H2O, give the masses of each of the products. 3). Hydrochloric acid (HCl gas dissolved in water) reacts with calcium carbonate in piece of limestone as follows: CaCO3 + 2HCI- CaCl2 + CO2 + H20. If 14.6 g of CO2 are produced in this reaction, what is the total mass of reactants and the total mass of the products?. a

Answer 1

Answer:

Explanation:

Problem 1:

Plants utilize light energy in the photosynthesis process to synthesize glucose, C₆H₁₂O₆, from CO₂ and H₂O by way of the reaction 6CO₂ +6H₂O → C₆H₁₂O₆ +60₂. How many grams of CO₂ are consumed in the production of 90 g of glucose?

 

Given parameters:

Mass of glucose = 10g

Unkown:

Mass of CO₂ consumed = ?

Solution:

 The reaction equation is given as:

                          6CO₂ + 6H₂O →  C₆H₁₂O₆ +60₂

• To solve this problem, the mole concept is a good approach. First work from the known to the unkown. The known is the given parameter which is the mass of glucose.

• Using the mass of glucose, estimate its number of moles and compare to that of the unknown carbon dioxide gas.

• Then find the mass from the obtained number of moles.

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Number of moles of C₆H₁₂O₆  = $\frac{mass of C₆H₁₂O₆}{molar mass of C₆H₁₂O₆}$

 given that atomic weight of:

               C = 12.01g/mol

               H = 1.01g/mol

               O = 16.00g/mol

  Molar mass of C₆H₁₂O₆ = (6 x 12.01) + (12 x 1.01) + (6 x 16) = 180g/mol

   Number of moles =  $\frac{90}{180}$ = 0.5mol

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from the reaction equation,

          1 mole of glucose is formed from 6 moles of CO₂

         0.5 mole of glucose would be formed from (6 x 0.5)mole = 3moles

##

    mass of CO₂   = number of moles x molar mass

   molar mass of CO₂ = 12 + 2(16) = 44g/mol

  mass of CO₂  = 3 x 44 = 132g of CO₂

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Problem 2:

For the reaction of 100.0 g of NaOH with Cl₂, 2NaOH + Cl₂  → NaClO + NaCI +H₂O, give the masses of each of the products.

Given parameters:

Mass of NaOH = 100g

Unknown:

mass of NaClo = ?

mass of NaCl = ?

mass of H₂O = ?

Solution:

Use the same procedure as highlighted in problem 1:

   equation of reaction:

             2NaOH + Cl₂  → NaClO + NaCI +H₂O

##

 Number of moles of NaOH = $\frac{mass of NaOH}{molar mass of NaOH}$

  Molar mass of NaOH:

     Atomic weight of Na = 23g/mol

     Atomic weight of O = 16g/mol

     Atomic weight of H = 1g/mol

 Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Number of moles of NaOH =  $\frac{100}{40}$ = 2.5mole

##

From the equation;

first product NaClO:  

           2 moles of NaOH produced 1 mole of NaClO

2.5 moles of NaOH will produce $\frac{2.5}{2}$mole = 1.25mole of NaClO

  mass of NaClO = number of moles x molar mass

    molar mass of NaClO = 23 + 35.5 + 16 = 74.5g/mol

  mass of NaClO = 1.25 x 74.5 = 93.13g of NaClO

Second product NaCl:

    2 moles of NaOH produced 1 mole of NaCl

   2.5 mole of NaOH will produce 1.25 mole of NaCl

mass of NaCl = number of moles x molar mass

  Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

   mass of NaCl = 1.25 x 58.5 = 73.13g of NaCl

Third product H₂O:

   2 moles of H₂O produced 1 mole of H₂O

   2.5 mole of H₂O will produce 1.25 mole of H₂O

 mass of H₂O = number of moles x molar mass

   molar mass of H₂O = 2(1) + 16 = 18g/mol

 mass of H₂O = 1.25 x 18 = 22.5g of H₂O

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Hydrochloric acid (HCl gas dissolved in water) reacts with calcium carbonate in piece of limestone as follows: CaCO₃ + 2HCI → CaCl₂ + CO₂ + H₂0. If 14.6 g of CO₂ are produced in this reaction, what is the total mass of reactants and the total mass of the products?

Given parameters:

Mass of CO₂ = 14.6g

Unkown:

Total mass of products = ?

Total mass of reactants =?

Solution:

Total mass of products = mass of CaCl₂ + mass of CO₂ + mass of H₂O

Total mas of reactants = mass of CaCO₃ + mass of HCl

Using the procedures highlighted in the first problem, solve for the number of moles of the known:

##

  Number of moles of CO₂ = $\frac{mass }{molar mass }$

   molar mass of CO₂ = 12 + 2(16) = 44g/mol

   Number of moles of CO₂ = $\frac{14.6}{44}$ = 0.33mole

##

number of moles of all species:

These species are in a ratio of one to one and the will have the same number of moles as that of the known:

  Number of moles of CaCl₂ = 0.33mole

  Number of moles of H₂O = 0.33mole

  Number of moles of CaCO₃ = 0.33mole

  For HCl:

     1 mole of CO₂ is produced from 2 moles of HCl

     0.33 mole of CO₂ will produce 2 x 0.33moles = 0.66moles of HCl

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Mass of each species:

Mass = number of moles x molar mass

   molar mass of CaCl₂ = 40 + 2(35.5) = 111g/mol

   molar mass of H₂O = 2(1 ) + 16 = 18g/mol

   molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol

   molar mass of HCl = 1 + 35.5 = 36.5g/mol

mass of CaCl₂ = 0.33 x 111 = 36.83g

mass of H₂O = 0.33 x 18 = 5.94g

mass of CaCO₃ = 0.33 x 100 = 33g

mass of HCl = 0.66 x 36.5 = 24.09g

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Total mass of reactants = 33g +  24.09g = 57.1g

Total mass of products = 36.83g  + 14.6g + 5.94g = 57.3g