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likoan [24]
3 years ago
7

C). If 0.5 mole of oxygen (,) reacted with hydrogen (H2), how many mole of water (H20) was produced

Chemistry
2 answers:
Harrizon [31]3 years ago
6 0

Answer:

1.0 mol of water will be produced.

Explanation:

Step 1: Data given

Number of moles of oxygen (O2) = 0.5 moles

Step 2: The balanced equation

2H2 + O2 → 2H20

Step 3: Calculate moles of water (H20)

For 2 moles of H2 consumed, we need 1 mol of O2 to produce 2 moles of water (H2O)

For 0.5 moles of oxygen (O2) we'll have 2*0.5 = 1.0 moles of water (H20) produced

1.0 mol of water will be produced.

son4ous [18]3 years ago
4 0

Answer:

1 mol of water is produced in those conditions.

Explanation:

The reaction to produce water between H₂ and O₂ is this:

2H₂  +  O₂  →  2H₂O

We don't have the amount of hydrogen, so we have to think that is in excess.

Let's work with oxygen.

Ratio is 1:2

For 0.5 mole of oxygen, I will make the double of moles of water.

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How did plastics become the material of choice for so many varied applications?
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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
AleksAgata [21]

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

Molar mass of zinc =65.38 g/mol

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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