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nata0808 [166]
3 years ago
11

Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily hcl, according to the reacti

on: mg(oh)2(aq) + 2 hcl(aq) → h2o(l) + mgcl2(aq) how much hcl in grams can be neutralized by 5.50 g of mg(oh)2 ? express your answer in grams to two decimal places.
Chemistry
2 answers:
Sonja [21]3 years ago
6 0

6.88 grams of hydrochloric acid (HCl) can be neutralized by 5.50 g of magnesium hydroxide, Mg(OH)2.

FURTHER EXPLANATION

This problem is an example of a simple stoichiometry problem. To determine the mass of hydrochloric acid neutralized, the following steps must be done:

  1. Balance the chemical equation.
  2. Convert the mass of magnesium hydroxide to moles using the molar mass.
  3. Use the stoichiometric ratio from the balanced chemical reaction to determine how many moles of hydrochloric acid are formed.
  4. Convert the moles of hydrochloric acid to mass using its molar mass.

Applying the steps above to the problem:

STEP 1: The balanced chemical equation is:

Mg(OH)₂(aq) + 2 HCl(aq) = 2 H₂O(l) + MgCl₂(aq)

STEP 2: The molar mass of Mg(OH)₂ is 58.3197 g/mol. The moles of Mg(OH)₂ can be calculated as follows:

moles \ Mg(OH)_{2} = 5.50 \ g Mg(OH)_{2} \times \frac{1 \ mol}{58.3197 \ g \ Mg(OH)_{2}}\\\\moles \ Mg(OH)_{2} = 0.09431 \ mol

STEP 3: The stoichiometric ratio of Mg(OH)₂ and HCl is 1:2. This can be used as a conversion factor to get the number of moles of HCl:

moles \ HCl = 0.09431 \ mol \ Mg(OH)_2 \times \frac{2 \ mol \ HCl}{1 \ mol \ Mg(OH)_{2}}\\\\moles \ HCl = 0.18862 \ mol

STEP 4: The molar mass of HCl is 36.46094 g/mol and this can be used to convert from moles HCl to grams HCl.

mass \ HCl \ = 0.18862 \ mol \ HCl \times \frac{36.46094 \ g}{1 \ mol \ HCl}\\mass \ HCl = 6.8788 \ g

Since the given has three significant figures, the final answer must also have 3 significant figures. Therefore,

\boxed {\boxed {mass \ HCl = 6.88 \ g}}

<h3>Learn More</h3>
  • Neutralization Reaction  brainly.com/question/4455839
  • Titration brainly.com/question/186765

<em>Keywords: stoichiometry, acid, base, moles</em>

Tema [17]3 years ago
5 0
For every 1 molecule of Magnesium hydroxide or Mg(OH)2 there will be 2 molecules of HCl neutralized.
If molar mass of magnesium hydroxide is 58.3197g/mol, the amount of mol in 5.50 g magnesium hydroxide should be: 5.50g/ (<span>58.3197g/mol)= 0.0943mol.
Then, the amount of HCl molecule neutralized would be: 2* </span>0.0943mol= 0.18861 mol

If molar mass of HCl is 36.46094 g/mol, the mass of the molecule would be: 0.18861 mol* 36.46094g/mol = 6.88grams
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Answer:

Explanation:

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Explanation:

The reaction expression is given as:

     Al(OH)₃  +   HNO₃   →    H₂O   +  Al(NO₃)₃  

To solve this problem, let us assign coefficient a,b,c and d to each specie;

          aAl(OH)₃  +   bHNO₃   →    cH₂O   +  dAl(NO₃)₃  

Conserving Al : a  = d

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 Multiply through by 3;

a  = 1, b  = 3, c = 3 and d  = 1

  Al(OH)₃  +   3HNO₃   →    3H₂O   +  Al(NO₃)₃  

7 0
3 years ago
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How many grams of Na2CO3 must be dissolved into 155g of water to create a solution with a molarity of 8.20 mol/kg
zhannawk [14.2K]

Answer:

135g Na2CO3

Explanation:

I'm going to assume you mean Molality which is mol solute/kg solvent

Molarity would be mol soute/ L solution

we know we have 155g of water which is .155 kg

essentially we have the equation:

mol/kg = 8.20

we substitute .155 in for kg and get:

mol/.155 = 8.20

Solving this  gives mol = 1.271

now we must convert to grams using the molar mass

Molar mass Na2CO3 = 106G/mol

so to cancel moles we multiply:

1.271mol x 106g/mol

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4 0
3 years ago
ObIel WiLll unt COl.. USSMS A certain chemical reaction releases 31.2 kJ/g of heat for each gram of reactant consumed. How can y
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Answer:

The expression to calculate the mass of the reactant is m = \frac{1.080kJ}{31.2kJ/g}

Explanation:

<em>The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant.</em> To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:

1,080J.\frac{1kJ}{10^{3}J } =1.080kJ

Then,

1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}

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3 years ago
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1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol (P^o_1) = 20.9 torr

Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

x_1 = mole fraction of 1‑Propanol

p_1=(0.540)\times (20.9torr)=11.3torr

and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
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Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
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