For the following chemical reaction, determine the amount of water in grams produced when 2.05 x 10 molecules of H react with 1.33 x 10 molecules of O to produce water 2H(g) + O(g) → 2HO(l)

For the following chemical reaction, determine the amount of water in grams produced when 2.05 x 10^23 molecules of H2 react with 1.33 x 10^23 molecules of O2 to produce water

Step 1: Data given

Number of H2 molecules = 2.05 *10^23 molecules

Number of O2 molecules = 1.33 * 10^23 molecules

Number of Avogadro = 6.022*10^23 / mol

Molar mass H2 = 2.02 g/mol

Molar mass O2 = 32.0 g/mol

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

2H2(g) + O2(g) → 2H2O(l)

Step 3: Calculate moles

Moles H2 = 2.05 * 10^23 / 6.022 * 10^23

Moles H2 = 0.340 moles

Step 4: Calculate moles O2

Moles O2 = 1.33 * 10^23 molecules / 6.022*10^23

Moles O2 = 0.221 moles

Step 5: Calculate limiting reactant

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

H2 is the limiting reactant. It will completely be consumed (0.340 moles).

O2 is in excess. There will react 0.340/2 = 0.170 moles

There will remain 0.221 - 0.170 = 0.051 moles

Step 6: Calculate moles H2O

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

For 0.340 moles we'll have 0.340 moles H2O

Step 7: Calculate mass H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 0.340 moles * 18.02 g/mol

Mass H2O = 6.13 grams

6.13 grams of H2O is produced