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3 years ago

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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A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f

Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

6 0

2 years ago

PLEASEEEEE HELPPPPPPPP MEEEEEEEEEEEEEEEE

NeTakaya

No, because boats and other mechanical vehicles can spill gas and oil into the freshwater. if the freshwater is scarce already, we should not contaminate it more by risking the gas and oil spills.

7 0

2 years ago

What is the percent magnesium in a sample made of 24g Mg and 70gCl?

seropon [69]

Answer:

Total percent of magnesium in sample = 25.5%

Explanation:

Given:

Mass of magnesium = 24 gram

Mass of chlorine = 70 gram

Find:

Total percent of magnesium in sample = ?

Computation:

Total mass of sample = Mass of magnesium + Mass of chlorine

Total mass of sample = 24 gram + 70 gram

Total mass of sample = 94 gram

Total percent of magnesium in sample = [Mass of magnesium / Total mass of sample]100

Total percent of magnesium in sample = [24/94]100

Total percent of magnesium in sample = [0.255]100

Total percent of magnesium in sample = 25.5%

5 0

3 years ago

The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.

adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

$\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}$

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

$\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol$

Now we have to calculate the energy required.

$Q=\frac{\Delta H}{n}$

where,

Q = energy required

$\Delta H$ = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

$Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ$

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

7 0

3 years ago

In the molecules below, areas that have a partial negative charge are pink and areas that have a partial positive charge are blu

Kaylis [27]

Answer:

Dipole-dipole interactions

Step-by-step explanation:

Each molecule consists of two different elements.

Thus, each molecule has permanent bond dipoles.

The dipoles do not cancel, so the attractive forces are dipole-dipole attractions.

"Covalent bonds" is wrong, because there are no bonds between the two molecules.

There are dipole-induced dipole and London dispersion forces, but they are much weaker than the dipole-dipole attractions.

4 0

3 years ago