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iren2701 [21]
3 years ago
15

What mass of ZnCI2 can be prepared from the reaction of 3.27 grams of zinc with 3.30 grams of HCI?

Chemistry
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer:

Zn. + 2 HCl ----------> ZnCl2. + H2

Explanation:

Mass of Zn in mole = 3.27/ 65.38= 0.050010.... mol

Mass of HCl in mole = 3.3/ 73= 0.044....mol

Hence, limiting reagent is HCl

Molar mass of ZnCl2= 136.38g

Let mass of ZnCl2 be x

x = 3.3*136.38÷ 73 = 6.165...

So, the mass of ZnCl2 is 6.17 g

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4 years ago
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answer

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sorry I do not know the answer, have a great day

7 0
3 years ago
A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
sweet-ann [11.9K]

Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT

Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

5 0
3 years ago
Zn + 2HCI --&gt; ZnCl2 + H2
Elena-2011 [213]
Moles= mass\ relative formula mass(Ar)
moles of zinc= 7.9/30= 0.263
so we have 0.263 moles of zinc, and you need twice the amount of chlorine so therefore 0.526moles of chlorine= 0.526x 17=8.942g of chlorine
i cba to work the rest out but the most reasonable answer is 0.24 mol however if you need to use working outs, use the formula i provided earlier
8 0
3 years ago
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