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makvit [3.9K]
3 years ago
14

How many moles of NAHCo3 are present in a 2.oog sample

Chemistry
1 answer:
Sav [38]3 years ago
6 0
Moles = grams/gfm

Na: 23
H: 1
C: 12
O: 48

gfm = 23+1+12+48 -> 84



So, 2g/84g/mol = 0.0238 moles of NaHCO3 
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Using stoichiometry, you predict that you should be able to use 314.0 g of Al to produce 1551 g of AlCi3. In your lab
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Answer:

90.26%

Explanation:

From the question given above, the following data were obtained:

Theoretical yield of AlCl₃ = 1551 g

Actual yield of AlCl₃ = 1400 g

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The percentage yield of the reaction can be obtained as follow:

Percentage yield = Actual yield / Theoretical yield × 100

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A reaction that produces 14.2 grams of a product and the theoretical yield of that product is 17.1 grams is true for the following statements :

The percent yield of the product is 83.0%

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<h3>Percentage Yield:</h3>

Percent yield is the percent ratio of actual yield to the theoretical yield.

Mathematically,

percent yield = actual yield / theoretical yield x 100%

actual yield = 14.2 grams

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percentage yield = 14.2 / 17.1 × 100%

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Therefore,

The percent yield of the product is 83.0%

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learn more on percentage yield here; brainly.com/question/4180677

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