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3 years ago

What Went Wrong – Balancing Chemical Equations

Chemistry

1 answer:

OLEGan [10]3 years ago

4 0

1-It has to be 3 Fe and not Fe3.
2-The oxygens aren't balanced

Balanced equation:
3Fe+4H2O---->Fe3O4+4H2

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What is the difference between long period and short period<br>.

Katarina [22]

The short run refers to a period of time short enough so that the amounts of at least one or more of the factors of production used by the firm cannot be changed. ... By the long run we mean a period of time long enough so that the amounts of all factors of production used by the firm can be changed.

8 0

2 years ago

How many electrons would be transferred in either a voltaic or electrolytic cell that uses the following half reactions

omeli [17]

Answer:

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

Explanation:

$Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V$

$Mg^{2+}(aq) + 2 e^- \rightarrow Mg (s),E^o = -2.37 V$

The substance having highest positive reduction $E^o$ potential will always get reduced and will undergo reduction reaction.

Reduction : cathode

$Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V$ ..[1]

Oxidation: anode

$Mg(s)\rightarrow Mg^{2+}(aq) + 2 e^-,E^o = 2.37 V$ ..[2]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

$E^o_{cell}=-0.036V-(-2.37 V)=2.334 V$

The overall reaction will be:

2 × [1] + 3 × [2] :

$2Fe^{3+} (aq) + 3Mg(s)+6e^-\rightarrow 2Fe (s)+3Mg^{2+}(aq)+6e^-$

Electrons on both sides will get cancelled :

$2Fe^{3+} (aq) + 3Mg(s)\rightarrow 2Fe (s)+3Mg^{2+}(aq)$

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

5 0

3 years ago

7 grams of oxygen gas is reacted with excess C4H8. How many grams of CO2 gas at STP are produced?

ladessa [460]

I think 14 are produced because if you go up by that you get it

4 0

2 years ago

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$