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DaniilM [7]
3 years ago
15

For the chemical reaction, 2HCl + Ca(OH)2 ➡️ CaCl2 + 2H2O what mass of calcium hydroxide in grams is needed to produce 3.63 mol

of water?
Chemistry
1 answer:
wlad13 [49]3 years ago
5 0

Answer:

2HCl (aq) + Ca

(

O

H

)

2

(aq) ---------> Ca

C

l

2

(ppt) + 2

H

2

O (aq)

let us calculate the number of moles , as per the chemical reactions;

2 moles of HCl solution reacts with one mole Calcium Hydroxide

Ca

(

O

H

)

2

One mole of HCl has mass : 36.5 g/mol, two moles of HCl will have mass, 73 g.

One mole of Ca

(

O

H

)

2

has mass 74.1 g

as per equation; 73 g of HCl reacts with 74.1 g of Ca

(

O

H

)

2

1g of HCl reacts with 74.1g / 73 of Ca

(

O

H

)

2

1g of HCl reacts with 1.015 g of Ca

(

O

H

)

2

NOW AS PER THE QUESTION MOLARITY AND VOLUME OF HYDROCHLORIC ACID IS GIVEN, IT CAN BE USED TO CALCULATE THE MASS OF HYDROCHLORIC ACID IN THE SOLUTION.

NUMBER OF MOLES of HCl ; Molarity of solution x Volume of Solution

# of moles of HCl = (0.40 mol/L ) x 350 mL

= (0.40 mol/L ) x 0.350 L = 0.14 mol

mass of HCl that makes 0.14 mol of HCl = # of moles x molar mass of HCl

mass of HCl = 0.14 mol x 36.5 g/ mol

mass of HCl = 5.11g

As per Stoichiometry , 1g of HCl reacts with 1.015 g of Ca

(

O

H

)

2

, 5.11g of HCl can react with 5.11 x 1.015 = 5.1865 g or 5.2 g of

Ca

(

O

H

)

2

.

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Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄

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:

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  • H₃PO₄: 1 mole
  • H₂O: 3 moles
  • Na₃PO₄: 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 H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?

moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}

moles of NaOH= 27 moles

But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.

<h3>Moles of Na₃PO₄ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?

moles of Na_{3}P O_{4} =\frac{6  moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}

<u><em>moles of Na₃PO₄= 2 moles</em></u>

Then, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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