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

There will be produced 18.02 grams of H2O

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

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