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fgiga [73]
3 years ago
7

For the following hypothetical reaction:

Chemistry
2 answers:
ycow [4]3 years ago
8 0

For the reaction:

3A +5B--->12C

Here 3 moles of A are reacting with 5 moles of B to produce 12 moles of C.

so, 1 mole of A will produce = 12÷3 moles of C

                                               =  4 moles of C

so 2 moles of A will produce = 4×2 moles of C

                                                 = 8 moles of C

so, 2 moles of A are reacting with excess of B to produce 8 moles of C.

so , the answer is 2 moles of A are reacting with excess of B to produce 8 moles of C.

nata0808 [166]3 years ago
3 0

Answer:

8 moles of compound C can you produce with 2 moles of compound A and excess compound B.

Explanation:

3A+5B → 12C

We are given with excess moles of compound B and 2 moles of compound A.

Moles of compound A = 2 moles

According to reaction, 3 moles of compound A gives 12 moles of compound C.

Then 2 moles of compound A will give:

\frac{12}{3}\times 2mol=8 mol of compound C.

8 moles of compound C can you produce with 2 moles of compound A and excess compound B.

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Limiting Reactants—————-
denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

  • Mg: 24.301;
  • O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}.

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}.

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. 2\times 0.270642 = 0.541284\;\text{mol} of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}.

Mass of 0.541284 moles of MgO:

m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}.

7 0
3 years ago
In science test are sometime clled
natita [175]

Answer:

In the scientific method, an experiment is an empirical procedure that arbitrates competing models or hypotheses. Researchers also use experimentation to test existing theories or new hypotheses to support or disprove them

Explanation:

Hope this helps

4 0
3 years ago
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Sunny_sXe [5.5K]

Answer:

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Explanation:

6 0
2 years ago
Which of the following is considered a STRONG electrolyte? C12H22O11 HC2H3O2 PbCl2 CH3OH NH4NO3? please explain
PIT_PIT [208]
I believe the correct answer from the choices listed above is the fifth option. Of the following , the strong electrolyte would be NH4NO3.  NH4NO3<span> is a salt and completely dissociates in aqueous solution. Hope this answers the question. Have a nice day.</span>
6 0
3 years ago
9. Suppose that 25.0 mL of a gas at 725 mm Hg and 20°C is converted to standard
storchak [24]

Answer:

V₂ = 22.23 mL

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 25 mL

Initial pressure = 725 mmHg (725/760 =0.954 atm)

Initial temperature = 20 °C (20 +273 = 293 K)

Final pressure = standard = 1 atm

Final temperature = standard = 273.15 K

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁  P₂

V₂ = 0.954 atm × 25 mL × 273.15 K / 293 K × 1 atm

V₂ =  6514.63 mL . atm . K  / 293 K . atm

V₂ = 22.23 mL

8 0
3 years ago
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