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zzz [600]
3 years ago
9

Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.3 g of methane is

mixed with 52.3 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.
Chemistry
1 answer:
Marrrta [24]3 years ago
6 0

Answer:

26 g

Explanation:

Write the balanced reaction first

CH4 + 2 O2 --> CO2 + 2 H2O

9.3g + 52.3g --> ? CO2

You must determine how much carbon dioxide can be made from each of the reactants. The maximum mass that can be made is the lower of the two.

From CH4:

9.3g CH4 (1molCH4/16.05gCH4) (1molCO2 / 1molCH4) (44.01g CH4 / 1molCO2) = 26 g

From O2:

52.3g O2 (1molO2/32gO2) (1molCO2/2molO2)(44.01g/1molCO2) = 36 g

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Answer:

5)HOCH2CH2OH

Explanation:

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The compounds with the highest amount of Hydrogen bond represents the one with the highest viscosity which is B) HOCH2CH2OH

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Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO
stiv31 [10]

Answer:

NO would form 65.7 g.

H₂O would form 59.13 g.

Explanation:

Given data:

Moles of NH₃ = 2.19

Moles of O₂ = 4.93

Mass of NO produced = ?

Mass of  produced H₂O = ?

Solution:

First of all we will write the balance chemical equation,

4NH₃ + 5O₂   →   4NO + 6H₂O

Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:

NH₃  :   NO                                   NH₃  :   H₂O

4     :    4                                          4    :      6

2.19   :    2.19                                 2.19  : 6/4 × 2.19 = 3.285 mol

Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:

O₂  :   NO                                               O₂ :   H₂O

5     :    4                                                  5     :    6

4.93   :   4/5×4.93 = 3.944 mol               4.93  : 6/5 × 4.93 = 5.916 mol

we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.

Mass of water = number of  moles × molar mass

Mass of water = 3.285 mol × 18 g/mol

Mass of water = 59.13 g

Mass of nitrogen monoxide  = number of  moles × molar mass

Mass of nitrogen monoxide = 2.19 mol × 30 g/mol

Mass of nitrogen monoxide = 65.7 g

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