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3 years ago

I WIll upvote for answer Balance the following reaction Fe^2 O ^3 +HCI+FeCI^3+H^2O

1 answer:

4 0

Let's balance step by step. We'll start with Iron (Fe)
Fe2O3 + HCl --> FeCl3 + H2O
We have 2 Fe on the left, and only one on the right, so we'll double the Fe on the right
Fe2O3 + HCl --> 2FeCl3 + H2O
Now we have six Cl on the right, and only one on the left, so we'll multiply the Cl by six on the left
Fe2O3 + 6HCl --> 2FeCl3 + H2O
Finally, we can balance the water, as we have 6 H and 3 O on the left, and 2 H and one O on the right, so we can triple the H2O on the right
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
The equation has been balanced.

You might be interested in

Find ΔHrxn for the following reaction: <br><br> 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)

ch4aika [34]

Answer:

ΔH°rxn = -827.5 kJ

Explanation:

Let's consider the following balanced equation.

2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)

We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]

ΔH°rxn = -827.5 kJ

5 0

3 years ago

Suppose you start with one liter of vinegar and repeatedly remove 0.08 L, replace with water, mix, and repeat. a. Find a formula

Bumek [7]

Answer:

0.92^n

Explanation:

Given that :

Initial amount of vinegar = 1 Litre

Number of litres removed repeatedly = 0.08 Litre

Since the amount removed each time is constant, then ;

Initial % = 100% = 100/100 = 1

. Using the relation :

Amount of vinegar in mixture :

Initial * (1 - amount removed / initial amount)^n

n = number of times repeated

1 * (1 - 0.08/1)^n

1 * (1 - 0.08)^n

1 * 0.92^n

Hence,

For nth removal,

Concentration will be :

0.92^n ; for n ≥ 1

7 0

2 years ago

Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 s

USPshnik [31]

<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

To calculate the moles of cadmium nitrate, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $MnSO_4$ = 0.796 M

Volume of $MnSO_4$ = 161 mL = 0.161 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol$

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

$MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)$

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = $\frac{1}{1}\times 0.13=0.13mol$ of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

$0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L$

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So, $0.183L=0.183\times 1000=183mL$

Hence, the volume of barium hydroxide is 183 mL.

7 0

3 years ago

Atoms like sodium and lithium often lose electrons, which makes them

zvonat [6]

More reactive than others

7 0

3 years ago

Read 2 more answers

Determine total H for bonds broken and formed, the overall change in H, and the final answer with units. Is it ENDOthermic or EX

Mrac [35]

E(Bonds broken) = 1371 kJ/mol reaction
E(Bonds formed) = 1852 kJ/mol reaction

ΔH = -481 kJ/mol.
The reaction is exothermic.

<h3>Explanation</h3>

2 H-H + O=O → 2 H-O-H

There are two moles of H-H bonds and one mole of O=O bonds in one mole of reactants. All of them will break in the reaction. That will absorb

E(Bonds broken) = 2 × 436 + 499 = 1371 kJ/mol reaction.
ΔH(Breaking bonds) = +1371 kJ/mol

Each mole of the reaction will form two moles of water molecules. Each mole of H₂O molecules have two moles O-H bonds. Two moles of the molecule will have four moles of O-H bonds. Forming all those bond will release

E(Bonds formed) = 2 × 2 × 463 = 1852 kJ/mol reaction.
ΔH(Forming bonds) = - 1852 kJ/mol

Heat of the reaction:

$\Delta H_{\text{rxn}} = \Delta H(\text{Breaking bonds}) + \Delta H(\text{Forming bonds})\\\phantom{ \Delta H_{\text{rxn}}} = +1371 + (-1852) \\\phantom{ \Delta H_{\text{rxn}}} = -481 \; \text{kJ} / \text{mol}$

$\Delta H_{\text{rxn}}$ is negative. As a result, the reaction is exothermic.

3 0

3 years ago