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s344n2d4d5 [400]
2 years ago
11

Help me with this question please

Chemistry
1 answer:
mash [69]2 years ago
6 0

<u>Answer:</u>

a. Oxidising agent: Cl₂

b. Reducing agent: NaBr

c. Oxidised: NaBr

d. Reduced: Cl₂

e. Oxidation numbers before reaction: Cl= 0, Na= +1, Br= -1

f. Oxidation numbers after reaction: Cl= -1, Na= +1, Br= 0

<u>Explanation:</u>

Oxidising agents reduces themselves, oxidising other elements/compounds.

Reducing agents oxidise themselves, reducing other elements/compounds.

Oxidation is the <u>loss</u> of electrons or an <u>increase</u> in oxidation number.

Reduction is the <u>gain</u> of electrons or <u>decrease</u> in oxidation number.

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Ferrophosphorus (Fe2P) reacts with pyrite (FeS2) producing iron(II) sulfide and a compound that is 27.87% P and 72.13% S by mass
k0ka [10]

Answer:

The molecular formula of the compound = P_4S_{10}

The empirical formula of the compound = P_2S_{5}

The balanced chemical equation for this reaction:

4Fe_2P+18FeS_2\rightarrow 26FeS+P_4S_{10}

Explanation:

Compound that is 27.87% P and 72.13% S by mass and has a molar mass of 444.56 g/mol.

Molar mass of compound = 444.56 g/mol

Number of phosphorus atom = x

Number of sulfur atom = y

Atomic  mass of phosphorus  31 g/mol

Atomic mass of sulfur = 32 g/mol

Percentage of element in compound :

=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100

Phosphorus :

27.87\%=\frac{x\times 31 g/mol}{444.56 g/mol}\times 100

x = 4

Sulfur :

72.13\%=\frac{y\times 32 g/mol}{444.56 g/mol}\times 100

y = 10

The molecular formula of the compound = P_4S_{10}

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

The empirical formula of the compound = P_2S_{5}

The balanced chemical equation for this reaction:

4Fe_2P+18FeS_2\rightarrow 26FeS+P_4S_{10}

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