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Veronika [31]
3 years ago
10

Balance the equation (NH4)3 PO4 +NA0H arrow Na3P04 +3NH3 +3H20.

Chemistry
2 answers:
mr_godi [17]3 years ago
8 0

Answer:

Explanation:

(NH4)3 PO4 +NaOH arrow Na3PO4 +3NH3 +3H2O

Start by seeing what happens with the Na. You need 3 on the left, so put a 3 in front of NaOH

(NH4)3 PO4 +3NaOH arrow Na3PO4 +3NH3 +3H2O  Next work with the nitrogens. YOu have 3 on the left and 3 on the right, so they are OK. Next Go to the stray oxygens.

You have 3 on left in (NaOH) and three on the right in 3H2O so they are fine as well. The last thing you should look at are hydrogens.

There are 12 + 3 on the left which is 15. There are 9 (in 3NH3) and 6 more in the water. They seem fine.

Why didn't I do something with the PO4^(-3)? The reason is a deliberately stayed away from them and balanced everything else. Since they were untouched with 1 on the left and 1 on the right, they are balanced.

Species      Na        H        O         N       PO4

Left             3          15        3         3          1

Right           3         15         3         3          1

Katena32 [7]3 years ago
3 0

Answer:

(NH₄)₃PO₄ + 3NaOH ⟶ Na₃PO₄ + 3NH₃ + 3H₂O

Explanation:

Your unbalanced equation is

(NH₄)₃PO₄ + NaOH ⟶ Na₃PO₄ + NH₃ + H₂O

A method that usually works for balancing by inspection is

  1. Balance all atoms other than O and H
  2. Balance O
  3. Balance H

1. Pick the most complicated-looking formula [(NH₄)₃PO₄].

Put a 1 in front of it.

<u>1</u>(NH₄)₃PO₄ + NaOH ⟶ Na₃PO₄ + NH₃ + H₂O

2. Balance N.

We have fixed 3N on the left. We need 3N on the right.

Put a 3 in front of NH₃.

<u>1</u>(NH₄)₃PO₄ + NaOH ⟶ Na₃PO₄ + <u>3</u>NH₃ + H₂O

3. Balance P.

We have fixed 1P on the left. We need 1P on the right.

Put a 1 in front of Na₃PO₄.

<u>1</u>(NH₄)₃PO₄ + NaOH ⟶ <u>1</u>Na₃PO₄ + <u>3</u>NH₃ + H₂O

4. Balance Na

We have fixed 3Na on the right. we need 3Na on the left.

Put a 3 in front of NaOH.

<u>1</u>(NH₄)₃PO₄ + <u>3</u>NaOH ⟶ <u>1</u>Na₃PO₄ + <u>3</u>NH₃ + H₂O

5. Balance O.

We have fixed 7O on the left and 4O on the right. We need three more O atoms on the right.

Put a 3 in front of H₂O.

<u>1</u>(NH₄)₃PO₄ + <u>3</u>NaOH ⟶ <u>1</u>Na₃PO₄ + <u>3</u>NH₃ + <u>3</u>H₂O

All species have a coefficient. The equation should now be balanced.

6. Check that all atoms are balanced

\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{N} & 3 & 3\\\text{H} & 15 & 15\\\text{P} & 1 & 1\\\text{O} & 7 & 7\\\text{Na} & 3 & 3\\\end{array}

The balanced equation is

(NH₄)₃PO₄ + 3NaOH ⟶ Na₃PO₄ + 3NH₃ + 3H₂O

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Given a daily mass of glucose:

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n_1 = {5.0\cdot 10^2 g}{180.156 g/mol} = 2.775 mol

From stoichiometry of this equation, moles of carbon dioxide can be found by multiplying this amount by 6:

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This is the mass of carbon dioxide per person per day. Multiply by the population and by the number of days to get the total mass:

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7 0
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Balance the following equations:
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Isobutyl propionate is the substance that provides the flavor for rum extract. combustion of a 1.152 g sample of this carbon-hyd
Ulleksa [173]

Answer;

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Solution;

Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)

Mass of carbon = 12/44 × 2.726 g

                           = 0.743455 g

Mass of Hydrogen  = 2/18 × 1.116 g

                            =  0.124 g

Mass of oxygen = 1.152 - (0.7435 + 0.124)

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Moles of carbon ;  0.7435/12 = 0.06196 moles

Moles of hydrogen; 0.124/1 = 0.124 moles

Moles of oxygen;  0.2845/16 = 0.01778 moles

Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778

         =  3.5 :  7.0 : 1

To make them whole numbers ; we multiply the ratios by 2  to get;

(3.5 :  7.0 : 1 )2 = 7 : 14 : 2

Thus, the empirical formula of Isobutyl propionate  is C7H14O2


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