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ELEN [110]
3 years ago
7

Calcium hydroxide, Ca(OH)2, is neutralized by hydrochloric acid, HCl, and the products are calcium chloride and water. You have

a sample of calcium hydroxide that you know contains 2.6x1024 hydroxide ions, based on an analysis of the sample. How many grams of HCl are required to react with the available calcium hydroxide?
Chemistry
1 answer:
padilas [110]3 years ago
7 0

n = m/Mr   (n= number of moles of a compound; m= mass of a compound; Mr = formula mass of a compound)

Rearranged: m = nxMr

Using the balanced equation for the reaction; <em>Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O   </em>we can see that there are 2 moles of OH- ions for every 2 moles of HCl and therefore we need to react the same number of moles of HCl as moles of OH- ions:

n, OH- ions = 2.6x10^24 mol

therefore:

n, HCl = 2.6x10^24 mol

Mr, HCl = 36.5 (use periodic table and the molecular masses of each element in the compound to find)

To calculate the mass of HCl needed, we multiply the number of moles of HCl and the Mr of HCl:

(2.6x10^24) x 36.5 = 9.49x10^25 g of HCl is required


You might be interested in
How many grams of hydrochloric acid are produced when 15.0 grams NaCl react with excess H2SO4 in the reaction
e-lub [12.9K]

Answer:

11.65 g

Explanation:

Amount of NaCl  = 15.0 grams

amount of H₂SO₄ = Excess

mass of hydrochloric acid (HCl) = ?

Solution:

To solve this problem first we will look for the reaction that NaCl react with  H₂SO₄

Reaction:

         2NaCl + H₂SO₄ --------> 2 HCl + Na₂SO₄

As the  H₂SO₄ is in excess so the amount of hydrochloric acid (HCl) depends on the amount of NaCl as it its act as limiting reactant.

Now if we look at the reaction

         2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

           1 mol                              2 mol  

Now convert moles to mass

Then

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

So,

            2NaCl           +      H₂SO₄     -------->       2HCl        +     Na₂SO₄

      2 mol (58.5 g/mol)                                    2 mol (36.5 g/mol)

              94 g                                                           73 g

So if we look at the reaction; 94 g of NaCl gives with 73 g of hydrochloric acid (HCl) in a this reaction, then how many grams of hydrochloric acid (HCl) will produce from 15.0 g of NaCl

For this apply unity formula

           94 g of NaCl  ≅ 73 g of HCl

           15.0 g of NaCl  ≅ X g of HCl

By Doing cross multiplication

           X g of HCl = 73 g x 15.0 g / 94 g

           X g of HCl = 11.65 g

11.65 g of hydrochloric acid (HCl) will produce by 15.0 g of NaCl

7 0
3 years ago
How do colloids that exist in nature affect people
mel-nik [20]

The exposure of a person to colloids majorly present in industrial area may results in the accumulation in the body tissue which causes dark deposit in the skin. They are not safe and is dangerous for health which may cause turning of skin blue permanently and it is life long.

Whereas, the colloid of silver nitrate is used to treat skin burns and ulcers but its overuse may result in adverse effect.

4 0
3 years ago
Tap water at room temperature has a pH of 7.2 and carbonate alkalinity of 200 mg/L (= 40 mmol/L). What is the concentration of b
UNO [17]

Answer:

the concentration of bicarbonate is <em>[HCO₃⁻] = 0,03996 M </em>and carbonate is <em>[CO₃²⁻] = 3,56x10⁻⁵ M.</em>

Explanation:

Carbonate-bicarbonate is:

HCO₃⁻ ⇄ CO₃²⁻ + H⁺ With pka = 10,25

Using Henderson-Hasselbalach formula:

pH = pka + log₁₀\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}

7,2 = 10,25 + log₁₀\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}

8,91x10⁻⁴ = \frac{[CO_{3}^{2-}]}{[HCO_{3}^-]} <em>(1)</em>

Also:

0,040 M = [CO₃²⁻] + [HCO₃⁻] <em>(2)</em>

Replacing (2) in 1:

<em>[HCO₃⁻] = 0,03996 M</em>

Thus:

<em>[CO₃²⁻] = 3,56x10⁻⁵ M</em>

I hope it helps.

4 0
3 years ago
Niven wants to calculate the mass of MgO that is produced by the burning of 28.0 g of Mg. What is the first step in Niven’s calc
Elanso [62]
I would say that Niven would have to calculate the atomic weights of Mg and O and then the total weight of MgO to get the percent of oxygen and then that way get a proportionate amount of the 28 grams of oxygen required to bond with the Mg and then add together the weight of Mg which would be 28.0 grams plus the weight of oxygen.
8 0
3 years ago
Read 2 more answers
En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello
MAXImum [283]

Answer:

a) m_{CH_4}=2630kg

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}

m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol

Con ello, los kilogramos de metano que cuestan 0,45 €:

44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0
4 years ago
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