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Leviafan [203]
3 years ago
7

How many moles of each reactant are needed to produce 3.60 x 10^2g ch3oh

Chemistry
1 answer:
zysi [14]3 years ago
3 0
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,

                                   CO  +  2 H₂    →    CH₃OH

Calculating Moles of CO:
                                       According to equation,

             32 g (1 mole) of CH₃OH is produced by  =  1 Mole of CO
So,
             3.60 × 10² g of CH₃OH is produced by  =  X Moles of CO

Solving for X,
                       X  =  (3.60 × 10² g × 1 Mole) ÷ 32 g

                       X  =  11.25 Moles of CO

Calculating Moles of H₂:
                                       According to equation,

             32 g (1 mole) of CH₃OH is produced by  =  2 Mole of H₂
So,
             3.60 × 10² g of CH₃OH is produced by  =  X Moles of H₂

Solving for X,
                       X  =  (3.60 × 10² g × 2 Mole) ÷ 32 g

                       X  =  22.5 Moles of H₂

Result:
            3.60 × 10² g of CH₃OH
is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
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Answer:

Balanced chemical equation:

C₃H₈ + 5O₂    →     4H₂O + 3CO₂

Explanation:

Chemical equation:

C₃H₈ + O₂    →     H₂O + CO₂

Balanced chemical equation:

C₃H₈ + 5O₂    →     4H₂O + 3CO₂

1st step:

Left side                               Right side

C = 3                                      C = 1

H = 8                                     H = 2

O = 2                                     O = 3

2nd step:

C₃H₈ + O₂    →     H₂O + 3CO₂

Left side                               Right side

C = 3                                      C = 1×3 = 3

H = 8                                    H = 2

O = 2                                     O = 6+1 = 7

3rd step:

C₃H₈ + O₂    →     4H₂O + 3CO₂

Left side                               Right side

C = 3                                      C = 1×3 = 3

H = 8                                     H = 2×4 = 8

O = 2                                     O = 6+4 = 10

4th step:

C₃H₈ + 5O₂    →     4H₂O + 3CO₂

Left side                               Right side

C = 3                                      C = 1×3 = 3

H = 8                                     H = 2×4 = 8

O = 2×5= 10                          O = 6+4 = 10

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