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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

3 years ago

How many grams of sodium metal are needed to make 29.3 grams of sodium chloride? given the reaction: 2na + cl2 → 2nacl?

Bess [88]

2Na + Cl2 → 2NaCl
46 g. ➡️ 117 g
x. ➡️ 29.3 g
$x = \frac{46 \times 29.3}{117} \\ x = 11.5 \: g$

5 0

3 years ago

Read 2 more answers

Classify methanal according to the position of the c=o group

seraphim [82]

Answer : Methanal also known as Formaldehyde $CH_{2}O$ is a chemical Aldehyde which contain ( -CHO) group.

Explanation :

In organic chemistry, a carbonyl group is a functional group which contain a carbon atom double-bonded to an oxygen atom i.e, ( C=O).

If carbonyl group is present in a compound then it can be a carboxylic (RCOOH), aldehyde (RCHO), ketone (RCOR'), ester ((RCOOR') or amide (RCONR'R") group.

Here are some functional groups naming according to the<em> IUPAC</em> rules and image also attached,

Carboxylic acid → (RCOOH) → ( name end in 'OIC ACID' )

Aldehyde → (RCOH) → ( name end in 'AL' )

Ketone → (RCOR') → ( name end in 'ONE' )

Ester → (RCOOR') → ( name end in 'ATE' )

Amide → (RCONR'R") → ( name end in 'AMIDE' )

In an aldehyde, atleast one hydrogen atom must be attached to the carbonyl carbon. For an aldehyde, remove ( -e) from alkane name and add ( -al) at the end of the compound.

Methanal is the IUPAC name for Formaldehyde.

3 0

3 years ago

A strand of DNA has the following string of bases:

Ganezh [65]

Answer:

ok so first of all try working hard then ill answer u

Explanation:

5 0

2 years ago

O<br> Plz answer this plz plz plz

saul85 [17]

Answer:

Q1. C

Q2 and Q3 are correct.

Explanation:

Since F=ma, and the force is a constant,

for the greatest acceleration, the mass of the ball must be the least.

Thus ball C has the greatest acceleration.

Let's check:

A) F=ma

a=F/m

a= F/68

B) a=F/72

C) a= F/64 (✓)

The smaller the denominator, the larger the value of a.

(Think: 1/2 >1/3)

3 0

3 years ago