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BARSIC [14]
2 years ago
7

How many grams of calcium chloride are needed to produce 10. 0 g of potassium chloride?

Chemistry
1 answer:
ozzi2 years ago
6 0

Answer:

<u>7.44 grams CaCl2 will produce 10.0 grams KCl.</u>

Explanation:

The equation is balanced:

I've repeated it here, with the elements corrected for their initial capital letter.

CaCl2( aq) K2CO3( aq) → 2KCl( aq) CaCO3( aq)

This equation tells us that 1 mole of CaCl2 will produce 2 moles of KCl.

If we want 10.0g of KCl, we need to convert that mass into moles KCl by dividing by the molar mass of KCl, which is 74.55 grams/mole.

 (10.0 grams KCl)/(74.55 grams/mole) = 0.1341 moles of KCl.

We know that we'll need half that amount of moles CaCl2, since the balanced equation says we'll get twice the moles KCl for every one mole CaCl2.

So we'll need (0.1341 moles KCl)*(1 mole CaCl2/2moles KCl) = 0.0671 moles CaCl2.

The molar mass of CaCl2 is 110.98 grams/mole.

(0.0671 moles CaCl2)*(110.98 grams/mole) = 7.44 grams CaCl2

<u>7.44 grams CaCl2 will produce 10.0 grams KCl.</u>

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4 years ago
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Please help quick
Rama09 [41]

Answer:

c = 0.898 J/g.°C

Explanation:

1) Given data:

Mass of water = 23.0 g

Initial temperature = 25.4°C

Final temperature = 42.8° C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Specific heat capacity of water is 4.18 J/g°C

ΔT = 42.8°C - 25.4°C

ΔT = 17.4°C

Q = 23.0 g ×  × 4.18 J/g°C × 17.4°C

Q = 1672.84 j

2) Given data:

Mass of metal = 120.7 g

Initial temperature = 90.5°C

Final temperature = 25.7 ° C

Heat released = 7020 J

Specific heat capacity of metal = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 25.7°C - 90.5°C

ΔT = -64.8°C

7020 J = 120.7 g ×  c ×  -64.8°C

7020 J = -7821.36 g.°C ×  c

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c = 0.898 J/g.°C

Negative sign shows heat is released.

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If a sample of 0.150 g of kcn is treated with an excess of hcl, calculate the amount of hcn formed in grams.
blondinia [14]
<h2> The amount of Hcn  formed in grams  is = 0.06223 grams</h2><h3> calculation</h3>

write the equation for reaction

that is;  KCn + HCl → KCl + HCn

 step 1:find the moles of KCn reacted

moles of KCn = mass of KCn/molar mass of KCn( 65.12 g/mol)

= 0.150 g/ 65.12 g/mol =2.303 x10^-3  moles

step 2: use the mole ratio between KCn :HCn  which is 1:1  to determine the moles of HCn   which is also  2.303 x 10^-3 moles

step 3: find the mass  of HCn =moles of HCn  x molar mass of HCn

= 2.303 x10^-3 x 27. 02 =0.06223 grams

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4 years ago
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