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shtirl [24]
2 years ago
14

Steve and Meg are doing an experiment where they need to make 37 grams of Fe2O3. How many grams of Fe would they need ?

Chemistry
1 answer:
Anna11 [10]2 years ago
5 0

Answer:

25.9g

Explanation:

Molecular mass of Fe₂O₃ = (56 * 2) + (16 * 3)

                                          = 112 + 48 = 160

Let's follow the unitary method now

In 160g Fe₂O₃ , there is 112 g Fe

In 1 g  Fe₂O₃ , there is 112 / 160 g Fe

In 37  Fe₂O₃ , we need 112 * 37 / 160

= 25.9 g Fe

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(3) Substance A is an element and substance Z is a compound.

Explanation;

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3 years ago
3.47 g of the hydrated "double salt", ammonium iron (II) sulfate hexahydrate, FeSO4(NH4)2SO4*6H2O was dissolved in 200. mL of wa
Ostrovityanka [42]

Answer:

1.4 × 10^-4 M

Explanation:

The balanced redox reaction equation is shown below;

5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O

Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol

Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles

Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M

Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M

Volume of Fe^2+ (VA)= 20.0 ml

Let the concentration of MnO4^- be CB (the unknown)

Volume of the MnO4^- (VB) = 12.6 ml

Let the number of moles of Fe^2+ be NA= 5 moles

Let the number of moles of MnO4^- be NB = 1 mole

From;

CAVA/CBVB = NA/NB

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CB= CAVANB/VBNA

CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5

CB = 1.4 × 10^-4 M

7 0
3 years ago
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