How far below sea level is the peak of the ridge?

Ammonia can be produced in the laboratory by heating ammonium chloride

AnnyKZ [126]

Answer:

Mass = 2.89 g

Explanation:

Given data:

Mass of NH₄Cl = 8.939 g

Mass of Ca(OH)₂ = 7.48 g

Mass of ammonia produced = ?

Solution:

2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O

Number of moles of NH₄Cl:

Number of moles = mass/molar mass

Number of moles = 8.939 g / 53.5 g/mol

Number of moles = 0.17 mol

Number of moles of Ca(OH)₂ :

Number of moles = mass/molar mass

Number of moles = 7.48 g / 74.1 g/mol

Number of moles = 0.10 mol

Now we will compare the moles of ammonia with both reactant.

NH₄Cl : NH₃

2 : 2

0.17 : 0.17

Ca(OH)₂ : NH₃

1 : 2

0.10 : 2/1×0.10 = 0.2 mol

Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.17 mol × 17 g/mol

Mass = 2.89 g

6 0

3 years ago

A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas

sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

CaCO3 + H2SO4 → CaSO4 + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4 × $\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}$ $\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}$