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4 years ago

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

Chemistry

1 answer:

Sav [38]4 years ago

5 0

Answer:

1 mol Al; 1 mol O2

Explanation:ol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

or the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

Express your answer aor the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

Express your answer as a chemical formula.

1 mol Al; 1 mol O2

4 mol Al; 2.5 mol O2

12 mol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

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B) Explain three ways by which water can be purified in the home

frosja888 [35]

Explanation:

Water can be purified from through...

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Chemicals used in water treatment...

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6 0

4 years ago

Calculate the concentration of IO−3 in a 1.65 mM Pb(NO3)2 solution saturated with Pb(IO3)2 . The Ksp of Pb(IO3)2 is 2.5×10−13 .

lara31 [8.8K]

Answer:

See explaination

Explanation:

1) Pb(NO3)2 => Pb2+ + 2 NO3-

[Pb2+] = [Pb(NO3)2] = 7.56 mM = 7.56 x 10-3 M

Pb(IO3)2 <=> Pb2+ + 2 IO3-

Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13

7.56 x 10-3 x [IO3-]2 = 2.5 x 10-13

[IO3-] = 5.75 x 10-6 M ≈ 5.8 x 10-6 M

(2) [Pb2+] = 1.7 x 10-6 M

Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13

1.7 x 10-6 x [IO3-]2 = 2.5 x 10-13

[IO3-] = 3.83 x 10-4 M

[IO3-]from Pb(IO3)2 = 2 x [Pb2+]

= 2 x 1.7 x 10-6 = 3.4 x 10-6 M

[IO3-]from NaIO3 = [IO3-] - [IO3-]from Pb(IO3)2

= 3.83 x 10-4 - 3.4 x 10-6

= 3.80 x 10-4 M

NaIO3 => Na+ + IO3-

[NaIO3] = [IO3-]from NaIO3

= 3.80 x 10-4 M ≈ 3.8 x 10-4 M

7 0

4 years ago

The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g)

Vladimir [108]

Answer:

$\Delta H=-29.7kJ$

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

$N_2+3H_2\rightarrow 2NH_3$

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

$n_{NH_3}^{by\ H_2}=1.96gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.647molNH_3\\\\ n_{NH_3}^{by\ N_2}=9.51gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.679molNH_3$

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

$\Delta H=0.647molNH_3*\frac{-45.9kJ}{1molNH_3}\\\\ \Delta H=-29.7kJ$

Best regards!

3 0

3 years ago

How many grams of Na2SO4 will be formed from 200.0g of NaOH

nikklg [1K]

The question is incomplete, the complete question is;

Using the following equation 2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq) how many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid

Answer:

355.1 g

Explanation:

The equation of the reaction is;

2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq)

We have been told that H2SO4 is in excess so NaOH is the limiting reactant. Therefore;

Number of moles in 200g of NaOH = 200g/40g/mol = 5 moles

So;

2 moles of NaOH yields 1 mole of Na2SO4

5 moles of NaOH will yield 5 * 1/2 = 2.5 moles of Na2SO4

Molar mass of Na2SO4 = 142.04 g/mol

Mass of Na2SO4= 2.5 moles * 142.04 g/mol = 355.1 g

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3 years ago

How many moles of propane

cupoosta [38]

Starting mass: 294
Mass ratio: 3/5

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2 years ago