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alexandr402 [8]
4 years ago
5

A chemist heats 100.0 g of FeSO4 x 7H2O in a crucible to drive off the water. If all the water is driven off, what is the mass o

f the remaining salt?
Chemistry
2 answers:
IrinaVladis [17]4 years ago
8 0

The correct answer is 54.6g

Ierofanga [76]4 years ago
4 0
FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)

M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol

m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)

m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)

m(FeSO₄)=151.9*100.0/278.0=54.6 g

m(FeSO₄)=54.6 g


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4 0
3 years ago
How many liters of ammonia (NH3), at 3.2 atm and 23C, must be used to produce of 2.65 grams of calcium hydride (CaH2). 6 Ca(s)
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Answer:

The answer to your question is    V = 0.32 L

Explanation:

Data

Volume of NH₃ = ?

P = 3.2 atm

T = 23°C

mass of CaH₂ = 2.65 g

Balanced chemical reaction

               6Ca  +  2NH₃   ⇒   3CaH₂  +  Ca₃N₂

Process

1.- Convert the mass of CaH₂ to moles

-Calculate the molar mass of CaH₂

 CaH₂ = 40 + 2 = 42 g

                             42 g ------------------ 1 mol

                              2.65 g --------------  x

                              x = (2.65 x 1)/42

                              x = 0.063 moles

2.- Calculate the moles of NH₃

                     2 moles of NH₃ --------------- 3 moles of CaH₂

                      x                        --------------- 0.063 moles

                                x = (0.063 x 2) / 3

                                x = 0.042 moles of NH₃

3.- Convert the °C to °K

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                      = 296°K

4.- Calculate the volume of NH₃

-Use the ideal gas law

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-Solve for V

                V = nRT / P

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                V = (0.042)(0.082)(296) / 3.2

-Simplification

               V = 1.019 / 3.2

-Result

               V = 0.32 L

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