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3 years ago

If 16 moles of al react with 3 moles of S8 how many moles of Al2 S3 will be formed

Chemistry

1 answer:

Gnom [1K]3 years ago

8 0

Answer:

8 moles

Explanation:

Al reacts with $S_8$ to produce $Al_2S_3$ as

$Al+S_8\rightarrow Al_2S_3$

The balanced chemical equation is

$16Al+3S_8\rightarrow 8Al_2S_3$

In the reaction, 16 moles of $Al$ react with 3 moles of $S_8$ to produce 8 moles of $Al_2S_3$ .

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How many protons does the neutral atom pictured have? A) 8 B) 18 C) 2 D) 20

Shalnov [3]

Answer:

b 18

Explanation:

3 0

3 years ago

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1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

Help!!!!

V125BC [204]

Answer:D and C sorry if im wrong

Explanation:

8 0

3 years ago

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S SIO₂ +3C Sic +20o Sic ? mass in grams С c-

seraphim [82]

Answer:

26.74g

Explanation:

The equation of the reaction is;

SIO₂ + 3C --> SiC +2CO

From the balanced equation, the relationship between SiC and C is;

3 mol of C produces 1 mol of SiC

Converting mol to mass using; Mass = moles * Molar mass

Mass of SiC = 1 mol * 40.11 g/mol = 40.11g

This means;

3 mol of C produces 40.11g of SiC

2 mol of C produces xg of SiC

3 = 40.11

2 = x

x = 2 * 40.11 / 3 = 26.74g

6 0

3 years ago

Write a balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air.

Lelechka [254]

Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

Explanation:

When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

$CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O$

Number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 2

Number of atoms present on product side are as follows.

C = 1
H = 2
O = 3

To balance this equation, multiply $O_{2}$ by 2 on reactant side. Also, multiply $H_{2}O$ by 2 on product side. Hence, the equation can be rewritten as follows.

$CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$

Now, the number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 4

Number of atoms present on product side are as follows.

C = 1
H = 4
O = 4

Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

4 0

3 years ago