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ddd [48]
3 years ago
11

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) When 25.0 g of Zn reacts, how many L of H2 gas are formed at STP?

Chemistry
2 answers:
ludmilkaskok [199]3 years ago
7 0
Answer to this is 8.56L
irga5000 [103]3 years ago
4 0

Answer : The volume of hydrogen gas formed at STP are, 8.557 liters

Explanation : Given,

Mass of zinc = 25.0 g

Molar mass of zinc = 65.38 g/mole

First we have to calculate the moles of zinc.

\text{Moles of }Zn=\frac{\text{Mass of }Zn}{\text{Molar mass of }Zn}=\frac{25g}{65.38g/mole}=0.382moles

Now we have to calculate the moles of hydrogen gas.

The given balanced chemical reaction is,

Zn(s)+2HCl(aq)\rightarrow H_2(g)+ZnCl_2(aq)

From the balanced chemical reaction, we conclude that

As, 1 mole of Zn react to give 1 mole of hydrogen gas

So, 0.382 mole of Zn react to give 0.382 mole of hydrogen gas

Now we have to calculate the volume of hydrogen gas.

At STP,

As, 1 mole of hydrogen gas contains 22.4 L volume of hydrogen gas

So, 0.382 mole of hydrogen gas contains 0.382\times 22.4=8.557L volume of hydrogen gas.

Therefore, the volume of hydrogen gas formed at STP are, 8.557 liters

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Zinc metal reacts with hydrochloric acid to produce zinc(II) chloride and hydrogen gas. How many liters of hydrogen gas will be
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Answer:

0.120 L of hydrogen gas will be produced

Explanation:

Step 1: Data given

Mass of zinc = 10.0 grams

Volume of hydrochloric acid = 23.8 mL

Molarity of hydrochloric acid = 0.45 M

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Step 2: The balanced equation

Zn + 2HCl → ZnCl2 + H2

Step 3: Calculate moles Zinc

Moles Zn = mass Zn / molar mass Zn

Moles Zn = 10.0 grams / 65.38 g/mol

Moles Zn  =  0.153 moles

Step 4: Calculate moles HCl

Moles HCl = molarity * volume

Moles HCl = 0.45 M * 0.0238 L

Moles HCl = 0.01071 moles

Step 5: Calculate limiting reactant

For 1 mol Zn, we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2

HCl is the limiting reactant. It will completely be consumed (0.01071 moles)

Zn is in excess. There will react 0.01071/2 = 0.005355 moles

There will remain 0.153 - 0.005355 = 0.147645 moles

Step 6: Calculate moles H2

For 1 mol Zn, we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2

For 0.01071 moles HCl we'll have 0.005355 moles H2

Step 7: Calculate volume H2

1 mol at STP = 22.4 L

0.005355 moles = 22.4 * 0.005355 = 0.120 L = 120 mL

0.120 L of hydrogen gas will be produced

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