Question

What mass of water can be obtained from 4.0 g of H2 and 16 g of O2?2 H2 + O2 ---> 2 H2O9 g18 g54 g36 g

Answer 1

Answer:

1.35g of H20 will be produced.

Step-by-step explanation:

1st) It is necessary to write the balanced chemical equation:

Now we know that 2 moles of H2 gas react with 1 mol of O2 gas to produce 2 moles of H2O.

2nd) With the molar mass of H2 (2g/mol) and O2 (32g/mol) we can find the limiting reactant. The limiting reactant is the one that will run out first and will not allow the reaction to continue:

• H2:

• O2:

Now we know that 4g of H2 needs 32g of O2 to react properly.

With a mathematical rule of three we can calculate the the grams of each compound using 9.5g of H2:

The 9.5g of H2 will need 76g of O2 to react properly, but we only have 1.2g of O2, so O2 is the limiting reactant.

3rd) Finally, using the 1.2g of the limiting reactant (O2), we can calculate the mass of H2O that will be produced:

So, 1.35g of H2O will be produced.

Step-by-step explanation:

Answer:

1.35g of H20 will be produced.

Step-by-step explanation:

1st) It is necessary to write the balanced chemical equation:

Now we know that 2 moles of H2 gas react with 1 mol of O2 gas to produce 2 moles of H2O.

2nd) With the molar mass of H2 (2g/mol) and O2 (32g/mol) we can find the limiting reactant. The limiting reactant is the one that will run out first and will not allow the reaction to continue:

• H2:

• O2:

Now we know that 4g of H2 needs 32g of O2 to react properly.

With a mathematical rule of three we can calculate the the grams of each compound using 9.5g of H2:

The 9.5g of H2 will need 76g of O2 to react properly, but we only have 1.2g of O2, so O2 is the limiting reactant.

3rd) Finally, using the 1.2g of the limiting reactant (O2), we can calculate the mass of H2O that will be produced:

So, 1.35g of H2O will be produced.

Answer 2

There will be produced 18.02 grams of H2O

Step-by-step explanation:

Step 1: Data given

Mass of H2 = 4.0 grams

Mass of O2 = 16 grams

Molar mass of O2 = 32 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

O2 + 2H2 → 2H2O

Step 3: Calculate moles of O2

Moles O2 = mass O2 / molar mass O2

Moles O2 = 16.0 grams / 32. g/mol

Moles O2 = 0.500 moles

Step 4: Calculate moles H2

Moles H2 = 4.0 grams / 2.02 g/mol

Moles H2 = 1.98 moles

Step 5: Calculate limting reactant

For 1 mol of O2 we need 2 moles of H2 to produce 2 moles of H2O

O2 is the limiting reactant. It will completely be consumed. (0.500 moles).

H2 is in excess. There will be react 2*0.500 = 1.00 moles

There will remain 1.98 - 1.00 = 0.98 moles

Step 6: Calculate moles of H2O

For 1 moles of O2 we need 2 moles of H2O

For 0.500 moles of O2 consumed, we'll produce 2*0.500 = 1.00 moles of H2O

Step 7: Calculate mass of H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 1.00 moles * 18.02 g/mol

Mass H2O = 18.02 grams

There will be produced 18.02 grams of H2O