If you have 1.90 mol of H_2 , how many grams of NH_3 can be produced?

N2 + 3H2 => 2NH3

mols H2 = 1.90
mols H2 x (2 mols NH3/3 mol H2) = 1.90 x 2/3 = ? mol NH3.
Then g = mols NH3 x molar mass NH3.

To determine the mass of NH3 that can be produced, we need to use the balanced chemical equation for the reaction between H2 and NH3. The balanced equation is:

3H2 + N2 -> 2NH3

This equation tells us that three moles of H2 react with one mole of N2 to produce two moles of NH3.

To solve the problem, we need to use stoichiometry, which is the proportional relationship between the amounts of reactants and products in a balanced chemical equation.

Given that we have 1.90 moles of H2, we can use the stoichiometry to find the moles of NH3 that can be produced. We can set up an equation like this:

(1.90 mol H2) x (2 mol NH3 / 3 mol H2) = x mol NH3

By multiplying 1.90 by 2/3, we can determine the number of moles of NH3 produced.

Next, we need to convert the moles of NH3 into grams. We can do this by using the molar mass of NH3, which is calculated by adding the atomic masses of nitrogen (N) and three hydrogens (H). The molar mass of NH3 is approximately 17.03 g/mol.

Using the calculated moles of NH3, we can find the mass of NH3 by multiplying the moles by the molar mass:

(x mol NH3) x (17.03 g NH3 / 1 mol NH3) = y g NH3

The resulting value of y will be the mass of NH3 that can be produced when starting with 1.90 mol of H2.