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3 years ago

ZnH2 + Cl2 → 2HCI + Zn

Chemistry

1 answer:

Ivahew [28]3 years ago

3 0

Answer:

2 moles H

Explanation:

Your tool of choice here will be the mole ratio that exists between zinc metal,

, and hydrochloric acid,

:HCl

, in the balanced chemical equation.

(s]

HCl

(aq]

→

ZnCl

2(aq]

2(g]

↑

⏐

You're dealing with a single replacement reaction in which zinc displaces the hydrogen from hydrochloric acid. The products of the reaction are aqueous zinc chloride and hydrogen gas.

Now, as you can see from the balance chemical equation, a

mole ratio exists between the two reactants.

This tells you that in order for the reaction to take place, you need to have twice as many moles of hydrochloric acid as you do of zinc metal.

http://people.springfield.k12.or.us/jim.tyser/chemcom/Resources/unit1ans.html

At the same time, you have a

mole ratio between hydrochloric acid and hydrogen gas.

This means that the reaction will always produce half as many moles of hydrogen gas as you have moles of hydrochloric acid.

Since you know that

moles of hydrochloric acid are taking part in the reaction, and assuming that you have enough zinc metal so that it doesn't act as a limiting reagent, you can say that the reaction will produce

moles HCl

⋅

1 mole H

moles HCl

2 moles H

Explanation:

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3 years ago

4. How many milligrams are in 5.25 x 10-13 kg? the “-13” is an exponent

rusak2 [61]

5. 25 x 10⁻⁷mg

Explanation:

This is mass conversion from mg to kg;

The kg is a quantity of mass used to measure the amount of matter in a substance.

Given mass = 5.25 x 10⁻¹³kg

The kilo- is a prefix that denotes 10³

therefore;

1000g = 1kilogram

the milli- is a prefix that denotes 10⁻⁻³

1000mg = 1g

Now that we know this, we can convert:

5.25 x 10⁻¹³kg x $\frac{1000g}{1kg} x \frac{1000mg}{1g}$ = 5. 25 x 10⁻¹³ x 10⁶mg

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3 years ago

Give me number of significant figures in the number 40

dangina [55]

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Which of the following interactions can contribute to the intrinsic binding energy during enzymatic catalysis? choose all that a

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3 years ago

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g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers