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nexus9112 [7]
3 years ago
13

During a science lab investigating chemical reactions, you mix 15 grams of baking soda with 20 grams of vinegar into a container

that has a mass of 55 grams. After the substances react what should the total mass of the products be after the reaction?
35


55


90


Cannot be determined
Chemistry
2 answers:
Marizza181 [45]3 years ago
7 0
55 as long as Ik I think it’s 55 but I’m Sorry if it’s wrong
Verdich [7]3 years ago
3 0
I think it is 55 but I could be wrongggggggg
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The by-product of the chlorination of an alkane is​
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The by-product of the chlorination of an alkane is​  <u>HCl</u>

Explanation:

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3 years ago
which one of the following groups are decomposers a. algae b. protist c. fungi d. green plants e. photosynthetic bacteria
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Explanation:

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Which substance is a compound iron or nitrogen oxide
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Read 2 more answers
A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?
Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles

Now we have to calculate the moles of Br_2

{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles

{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles

The balanced chemical reaction is,

2Na(s)+Br_2(g)\rightarrow 2NaBr

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = \frac{2}{1}\times 0.189=0.378 moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = \frac{2}{1}\times 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%

Therefore, the percent yield is, 93.4%

3 0
3 years ago
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