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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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(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

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3 years ago
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The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the rea
sineoko [7]
We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula: 
Kc   = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 =  0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

4 0
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