M(H2)=3г m(O2)=64г m(H2O)=?

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вопрос

To find the mass of water (H2O) produced from the reaction of hydrogen (H2) and oxygen (O2), we need to use the balanced chemical equation for the reaction:

2H2 + O2 -> 2H2O

From the equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

First, calculate the number of moles of hydrogen and oxygen given the masses provided:

Molar mass of H2 = 2g/mol
Moles of H2 = 3g / 2g/mol = 1.5 mol

Molar mass of O2 = 32g/mol
Moles of O2 = 64g / 32g/mol = 2 mol

Next, determine the limiting reactant in the reaction. One mole of O2 can react with 2 moles of H2 to produce 2 moles of water. Since there are only 1.5 moles of H2 available, it is the limiting reactant.

Now, we can calculate the amount of water produced using the moles of H2:

Moles of H2O = 1.5 mol * (2 mol H2O / 2 mol H2) = 1.5 mol

Now, calculate the mass of water produced:

Molar mass of H2O = 18g/mol
Mass of H2O = 1.5 mol * 18g/mol = 27g

Therefore, the mass of water produced from the reaction is 27 grams.