What will happen if you add more solute to a saturated solution?

How many moles of AlCl3 are formed by 1.53 moles of CuCl2?

jolli1 [7]

Answer:

1.02 moles of AlCl3

Explanation:

Consider the balanced reaction equation shown below;

2Al(s) + 3CuCl2(aq) -------> 2AlCl3 (aq) + 3Cu(s)

This is the balanced reaction equation for the reaction going on above. Recall that the first step in solving any problem is to accurately put down the chemical reaction equation. This balanced reaction equation always serves a reliable guide in solving the problem at hand.

Given that;

3 moles of CuCl2 yields 2 moles of AlCl3

1.53 moles of CuCl2 yields 1.53 × 2 / 3 =1.02 moles of AlCl3

5 0

2 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

2 years ago

If the molar mass of NaCl is 58.44 g/mol, what amount of NaCl (in mol) is contained in 250.0 mg of NaCl?

4vir4ik [10]

Answer: 0.004 moles

Explanation:

Amount of substance= mass/ molar mass

Mass= 250mg or 0.25g

Molar mass= 58.44gmol-1

Amount= 0.25/58.44

= 0.004 moles

5 0

2 years ago

You are a researcher for a golf club manufacturer. You are given two identical looking cubes of a metal alloy. You are informed

Rashid [163]

Answer:

C.Melt both cubes and look for a broader range of melting temperatures. The one that melts over a broader range of temperatures is the amorphous solid.

Explanation:

Amorphous solids is one that do not have a fixed melting points but melt over a wide range of temperature due to the irregular shape hence its name. Contrariwise crystalline solids, have a fixed and sharp melting point.

This comes in handy to solve the riddle. We can characterise the pair with the melting point property.

7 0

3 years ago

At a cost of $1600/oz, how much would you have to pay for a solid cubic foot of gold?

Alex777 [14]

We'll begin by calculating the mass in ounce (oz) of a cube foot (ft³) of gold. This can be obtained as follow:

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Density of gold = 19298 oz/ft³

Volume of gold = 1 ft³

Density = mass /volume

19298 = mass / 1

Finally, we shall determine the cost of 19298 oz of gold. This can be obtained as follow:

1 oz = $ 1600

Therefore,

19298 oz = 19298 × 1600

19298 oz = $ 30876800

Therefore, a solid cube foot of gold (i.e 19298 oz) will cost $ 30876800

Learn more: brainly.com/question/15407624

4 0

2 years ago