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3 years ago

7. Calculate moles of H2 needed to form 2.25 moles HCl in the reaction H2 + Cl2==>2HCl

Chemistry

1 answer:

mart [117]3 years ago

8 0

Answer:

1.125 moles

Explanation:

2mole of HCl produced 1mole of H2

2.25moles of HCl will produce x moles

cross multiply

2x=™1×2.25

x= 2.25÷2

x=1.125mole

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A large balloon is initially filled to a volume of 25.0 L at 353 K and a pressure of 2575 mm Hg. What volume of gas will the bal

bija089 [108]

Answer:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 25.0 L

V₂ = ?

P₁ = 2575 mm Hg

Also, P (atm) = P (mm Hg) / 760

P₁ = 2575 / 760 atm = 3.39 atm

P₂ = 1.35 atm

T₁ = 353 K

T₂ = 253 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}$

$\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}$

Solving for V₂ , we get:

V₂ = 45.0 L

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

4 0

4 years ago

A substance formed when two or more chemical elements are chemically bonded together?

Ronch [10]

Answer: a piece of pure matter which we can see from our naked eye is termed as a substance

Explanation:

every substance has physical and chemical properties like;

diamond ,water ,pure sugar ,table salt e.t.c

an element is a substance made up of same type of atoms having same atomic number[protons],and cannot be easily decomposed in to a single substance by ordinary chemical reaction example,.....

gold,copper,zinc and some are in liquid form like mercury,bromine and some are in gases for example, hydrogen ,oxygen, nitrogen

6 0

3 years ago

Which statement defines dynamic equilibrium? (5 points)

Shalnov [3]

Answer:

1. A state of balance in which the rates of the forward and reverse reactions are equal.

Explanation:

A dynamic equilibrium is like a cycle, the reactants change to products, but the products also change to reactants keeping the amount of each constant.

2. A state of balance in which the forward reaction stops but reverse reaction continues.

In this statement there isnt a equilibrium. The products will change to reactants until the reaction stops.

3. A state of balance in which the forward reaction continues but reverse reaction stops.

Here the reactants will change to products until the reaction stops.

4. A state of balance in which the forward and reverse reactions stop.

In this case the reaction has stopped.

4 0

3 years ago

which statement best describes the equation caco3 2hcl → cacl2 co2 h2o? caco3 is a reactant; it is present before the reaction o

fomenos

Answer: $CaCO_3$ is a reactant; it is present before the reaction occurs.

Explanation:

In a chemical reaction the chemical formulas written before the arrow are described as reactants as they react together to form products which are written after the arrow.

$CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$

Thus $CaCO_3$ and HCl are reactants here whereas $CO_2$ , $CaCl_2$ and $H_2O$ are products.

7 0

3 years ago

Read 2 more answers

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

3 years ago