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Mumz [18]
3 years ago
9

ribose is an important sugar that is found in DNA and RNA.Ribose has a molar mass of 150g/mol,and a percent composition of 40.0%

C,6.67%H,and 53.5%O.What is the molecular formula for ribose
Chemistry
1 answer:
DENIUS [597]3 years ago
3 0
40.0 gC x 1 mol C/12.0 gC=3.33 moles of C ----- divide by smallest
6.67 gH x 1 mol H/ 1.0 gH= 6.7 moles of H ------ CH2O= (12 + 2 + 16)= 30g
53.3 gO x 1mol O/ 15.0 gO=3.33 moles of O ----- 150g/30g= 5x
=C5H10O5
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An alkaline battery produces electrical energy according to the following equation.
Gnesinka [82]

Answer:

Part a: limiting reactant MnO₂

Part b: 12.43 g of Zn(OH)₂

Explanation:

Part a

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Limiting reactant = ?

Given Reaction

        Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

To determine the limiting reactant first we will look at the reaction

         Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

          1 mol     2 mol

Convert moles to mass

molar mass of Zn = 65.4 g/mol

Mass of MnO₂ = 55 + 2 (16) = 55 + 32 = 87 g/mol

So,

       Zn(s)          +      2 MnO₂(s)    +    H₂O(l) ----------> Zn(OH)₂(s) + Mn₂O₃(s)

1 mol (65.4 g/mol)    2 mol (87 g/mol)

       65 g                        174 g

So its clear from the reaction that 65 g Zn react with 174 g of MnO₂.

now if we look at the given amounts the amount MnO₂ is less then the amount of Zn but in actual calculation amount of MnO₂ is more then amount of zinc.

So, for MnO₂ if we calculate the needed amount of zinc

So apply unity formula

           65 g Zn react ≅ 174 g of MnO₂

            X g of Zn ≅ 21.5 g of MnO₂

Do cross multiplication

           X g Zn react = 65 g x 21.5 g / 174 g

           X g of Zn ≅ 8.032 g

So, 8.032 g of zinc will react out of 37.0 grams. the remaining will be in excess.

So MnO₂ will be consumed completely an it will be limiting reactant.

____________

part b

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Mass of Zn(OH)₂  produced = ?

Given Reaction

        Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

Solution:

As from the part A we come to know that MnO₂ is limiting reactant, so the amount of Zn(OH)₂ will depend on the amount of MnO₂.

So first we convert mass of MnO₂ to moles

Formula Used

        no. of moles = mass in grams / molar mass

Mass of MnO₂ = 55 + 2 (16)

Mass of MnO₂ = 55 + 32 = 87 g/mol

Put values in the above equation

        no. of moles = 21.5 g / 87 g/mol

        no. of moles = 0.25 moles

Now,

Look at the reaction

         Zn(s)  +  2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

                        2 mol                                       1 mol  

So its clear from the reaction that 2 mole of MnO₂ gives 1 mole of Zn(OH)₂

then how many moles of Zn(OH)₂ will be produce by 0.25 moles of MnO₂

So.

apply unity formula

         2 mol of MnO₂ ≅ 1 mole of Zn(OH)₂

            0.25 moles of MnO₂  ≅ X mole of Zn(OH)₂

Do cross multiplication

           X mole of Zn(OH)₂ = 1 mole x 0.25 mol x  / 2 mol

         X mole of Zn(OH)₂  ≅ 0.125

Now Conver moles of  Zn(OH)₂  to mass

Formula used

         mass in grams = no. of moles x molar mass

Molar mass of  Zn(OH)₂

Molar mass of  Zn(OH)₂ = 65.4 + 2 (16 + 1)

Molar mass of  Zn(OH)₂ = 65.4 + 2 (17)

Molar mass of  Zn(OH)₂ = 65.4 + 34 = 99.4 g/mol

Put values in above equation

        mass in grams = 0.125 mol x 99.4 g/mol  

        mass in grams = 12.43 g

So,

12.43 g of Zn(OH)₂  will be produce.

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