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RideAnS [48]
3 years ago
5

Consider the reaction: 3Co2+(aq) + 6NO3¯(aq) + 6Na+(aq) + 2PO43¯(aq) â Co3(PO4)2(s) + 6Na+(aq) + 6NO3¯(aq) Identify the net i

onic equation for this reaction.A. 3CO2(aq) + 6NO3(aq) + 6Na+(aq) + 2PO43-(aq) â Co3(PO4)2(s) + 6NaNO3(aq). B. Na+(aq) + NO3-(aq) â NaNO3(aq).C. 3CO2(aq) + NO3(aq) + Na*(aq) + 2PO43-(aq) â Co3(PO4)2(s) + NaNO3(aq).D. 3C02(aq) + 2PO43-(ag) â CO3(PO4)2(s).E. None of these choices are correct.
Chemistry
1 answer:
irina [24]3 years ago
3 0

Answer:

D. 3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

Explanation:

The complete ionic equation includes all the ions and the insoluble species. Let's consider the following complete ionic equation.

3 Co²⁺(aq) + 6 NO₃⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 NO₃⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble species. The corresponding net ionic equation is:

3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

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The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.
Kitty [74]

Answer:

a) The theoretical yield is 408.45g of BaSO_{4}

b) Percent yield = \frac{realyield}{408.45g}*100

Explanation:

1. First determine the numer of moles of BaCl_{2} and Na_{2}SO_{4}.

Molarity is expressed as:

M=\frac{molessolute}{Lsolution}

- For the BaCl_{2}

M=\frac{1.75molesBaCl_{2}}{1Lsolution}

Therefore there are 1.75 moles of BaCl_{2}

- For the Na_{2}SO_{4}

M=\frac{2.0moles[tex]Na_{2}SO_{4}}{1Lsolution}[/tex]

Therefore there are 2.0 moles of Na_{2}SO_{4}

2. Write the balanced chemical equation for the synthesis of the barium white pigment, BaSO_{4}:

BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the BaCl_{2}:

\frac{1.75}{1}=1.75

- For the Na_{2}SO_{4}:

\frac{2.0}{1}=2.0

As the BaCl_{2} is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = \frac{realyield}{theoreticalyield}*100

Percent yield = \frac{realyield}{408.45g}*100

The real yield is the quantity of barium white pigment you obtained in the laboratory.

7 0
3 years ago
PLS HELP CHEM
Bess [88]

Answer:

Use the formula PV=nRT to get the temperature

7 0
2 years ago
If the volume of an object increases but the mass stays the same, which will happen to the density?
nikklg [1K]
We all know, Density = Mass/ volume. When volume increases and the mass remains the same , the density will decrease considerably.
5 0
3 years ago
Read 2 more answers
What is the molarity of a solution containing 12.0 g of NaOH in 250.0 mL of solution?
riadik2000 [5.3K]

Answer:

1.196 M NaOH

Explanation:

Molarity = moles/Volume (L)

moles NaOH = mass NaOH/MM NaOH = 12/40.01 = 0.299 moles NaOH

Volume solution = 250 mL = 0.250L

M = 0.299/0.250=1.196 M NaOH

3 0
3 years ago
a) Provide equation of K of this reaction, use symbol " ^ " for exponents. That means 1000 = 10^3 and 1/100 is 10^(-2). b) How m
stealth61 [152]

Answer and Explanation:

a. The equation of K of this reaction is shown below:-

3 A + 5 B + 4 C↔5 D + 7 E + F

K = \frac{(D)^5 (E)^7 (F)}{(A)^3 (B)^5 (C)^4}

b. The moles of compound F is shown below:-

3 A + 5 B + 4 C↔5 D + 7 E + F

               2 moles

Now, the mole of produced is

= \frac{1}{4} \times \ moles\ of\ c

Now, we will the value of c by using the above equation

= \frac{1}{4} \times 2

After solving the above equation we will get

0.5 moles

5 0
3 years ago
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