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Sauron [17]
3 years ago
8

2NH3 + MgSO4 + 2H20-→ Mg(OH)2 + (NH4)2SO4

Chemistry
2 answers:
gayaneshka [121]3 years ago
7 0
Chemistry msg for 78 mgSO 5.4 answer is NH3
Vlada [557]3 years ago
7 0

Answer:

Answer is NH3 is the limiting reactant

Explanation:

There is excess MgSO4

4.6 moles - 2.7 moles = 1.9 moles left over of MgSO4 as excess

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Answer: A physical trait is a defining aspect or feature of your body or other subject that you can see visually.

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3 years ago
2. The empirical formula of a molecule is CH2O. In an experiment, the molar mass of the molecule was determined to be 360.3 g/mo
Vanyuwa [196]

Answer:

The answer to your question is:

2.- C₁₂ H₂₄ O₁₂

Explanation:

2.-

Data

CH2O

molar mass = 360.3 g/mol

Molar mass of CH2O = 12 + 2 + 16 = 30g

Divide molar mass given by molar mass obtain

                   

                           x = 360.3/30

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Finally

                      C₁₂ H₂₄ O₁₂

Molar mass = (12 x 12) + (24 x 1) + (16 x 12) = 144 + 24 + 192 = 360 g

3.- First we need to write the complete equation of the reaction and balanced it.

Then, we need to convert the mass given to moles of each compound.

After that, we need used rule of three calculate the amount of products based on the moles of reactants given.

Finally, convert the moles to grams.

4.-

a.- It is a relation between the mass of product obtain in an experiment and the mass of a product obtain theoretically times 100.

b.-

35 g of Mg reacted with excess O2

percent yield = 90%

Actual yield = ?

Formula

Percent yield = (actual yield/theoretical yield) x 100

Equation  

                       2Mg  + O2 ⇒ 2MgO

                      48.62 g of Mg ----------------- 80.62 g of MgO

                      35g                  ------------------  x

                     x = 58 g of MgO     (Theoretical yield)

Theoretical yield = 58 g of MgO

Actual yield = percent yield x theoretical yield / 100

                    = 90 x 58 / 100

                   = 52. 23 g

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where

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2.

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where

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Substituting,

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Answer:

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