Is phosphorus tribromide a covalent compounds or ionic compounds

velikii [3]

Taking into account the reaction stoichiometry and definition of limiting reactant, AgNO₃ will be the limiting reagent.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cu: 1 mole
AgNO₃: 2 moles
Ag: 2 moles
Cu(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Cu reacts with 2 moles of AgNO₃, 1.8 moles of Cu reacts with how many moles of AgNO₃?

$amount of moles of AgNO_{3} =\frac{1.8 moles of Cux2 moles of AgNO_{3} }{1 mole of Cu}$

<u><em>amount of moles of AgNO₃= 3.6 moles</em></u>

But 3.6 moles of AgNO₃ are not available, 2 moles are available. Since you have less moles than you need to react with 1.8 moles of Cu, AgNO₃ will be the limiting reagent.

<h3>Summary</h3>

In summary, AgNO₃ will be the limiting reagent.

Learn more about the reaction stoichiometry:

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8 0

1 year ago

What mass of copper is required to replace silver from 4.00g of silver nitrate dissolved in water?

larisa [96]

The balanced chemical reaction is written as:

<span>Cu +2AgNO3 → Cu(NO3)2 + 2Ag
</span>
We are given the amount of silver nitrate to be used for the reaction. This value will be the starting point of our calculations. It is as follows:

4.00 g AgNO3 ( 1 mol / 169.87 g ) ( 1 mol Cu / 2 mol AgNO3 ) ( 63.456 g / 1 mol ) = 0.747 g Cu

3 0

3 years ago

Pure water is an electrolytic solution. <br><br><br> True <br> False

dmitriy555 [2]

The answer to this is false.

6 0

3 years ago

A sample that contains only SrCO3 and BaCO3 weighs 0.800 g. When it is dissolved in excess acid, 0.211 g car- bon dioxide is lib

MrRissso [65]

Answer:

53.9%

Explanation:

1 mole of BaCO₃ yields 1 mole of CO₂,

1 mole of SrCO₃ yields 1 mole of CO₂

m₁ = mass of BaCO₃

m₂ = mass SrCO₃

molar mass of SrCO₃ = 147.63 g/mol

molar mass of BaCO₃ = 197.34 g/mol

molar mass of CO₂ = 44.01 g/mol

mole of CO₂ in 0.211 g = 0.211 g / 44.01 = 0.00479

mole of BaCO₃ = m₁ / 197.34

mole of SrCO₃ = m₂ / 147.63

mole of BaCO₃ + mole of SrCO₃ = 0.00479

(m₁ / 197.34) + (m₂ / 147.63) = 0.00479

147.63 m₁ + 197.34 m₂ = 139.55

m₁ + m₂ = 0.8

m₁ = 0.8 - m₂

147.63 (0.8 - m₂) + 197.34 m₂ = 139.55

118.104 - 147.63 m₂ + 197.34 m₂ = 139.55

49.71 m₂ = 139.55 - 118.104 = 21.446

m₂ = 21.446 / 49.71 = 0.431

the percentage of m₂ ( SrCO₃ ) in the sample = 0.431 / 0.8 = 0.539 × 100 = 53.9%

5 0

3 years ago

A sample was prepared by mixing 18. ml of 3.00 x 10^-3 m crystal violet (cv) with 2.00 ml of 0.250 m naoh. calculate the resulti

Aleks [24]

Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.

Explanation :

Step 1 : Find moles of crystal violet and NaOH.

The molarity formula is

$Molarity = \frac{mol}{L}$

Molarity of crystal violet = $3.00 \times 10^{-3} = \frac{mol (CrystalViolet)}{L}$

The volume of crystal violet solution is 18 mL which is 0.018 L.

Moles of crystal violet = $3.00 \times 10^{-3} \times 0.018 = 5.4 \times 10^{-5}$

Moles of crystal violet = 5.4 x 10⁻⁵

Moles of NaOH = $Molarity \times L = 0.250 \times 0.00200 = 5.00 \times 10^{-4}$

Moles of NaOH = 5.00 x 10⁻⁴

Step 2 : Find total volume of the solution

The total volume of the solution after mixing NaOH and crystal violet is

0.018 L + 0.00200 = 0.020 L

Step 3 : Use molarity formula to find final concentrations

Molarity of crystal violet = $\frac{mol(CrystalViolet)}{Total Volume(L) } = \frac{5.4 \times 10^{-5}}{0.020} = 2.7 \times 10^{-3}$

Final concentration of CV = 0.0027 M

Molarity of NaOH= $\frac{mol(NaOH)}{Total Volume(L) } = \frac{5.00 \times 10^{-4}}{0.020} = 0.025 \times 10^{-3}$

NaOH is a strong base and dissociates completely as follows.

$NaOH (aq) \rightarrow Na^{+} (aq) + OH^{-} (aq)$

The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.

Concentration of OH⁻ = 0.025 M

8 0

3 years ago