Its 17g. in the reaction
N2+3H2-->2NH3

14g of N2 reacts with 6g of H2 to form 34g of NH3,thus 14g would give 17g by reacting with 3g of H2.
here N2 is limiting reagent.

Solution:

The balanced chemical equation for the reaction of nitrogen gas and hydrogen gas to produce ammonia is as follows:

N2(g) + 3H2(g) → 2NH3(g)

From the equation, it can be observed that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia gas. Therefore, the ratio of nitrogen gas to hydrogen gas in the reaction is 1:3.

Given that 14g of nitrogen gas is mixed with 6g of hydrogen gas, we can calculate the maximum mass of ammonia that can be produced in the reaction as follows:

1. Determine the limiting reactant:

The limiting reactant is the reactant that is completely consumed in the reaction, thus limiting the amount of product that can be formed. To determine the limiting reactant, we need to compare the amount of each reactant with their respective stoichiometric coefficients in the balanced chemical equation.

Moles of nitrogen gas = Mass / molar mass = 14g / 28g/mol = 0.5 mol

Moles of hydrogen gas = Mass / molar mass = 6g / 2g/mol = 3 mol

Since the ratio of nitrogen gas to hydrogen gas in the reaction is 1:3, we can see that there is excess hydrogen gas available for the reaction. Therefore, nitrogen gas is the limiting reactant.

2. Calculate the maximum mass of ammonia produced:

Using the stoichiometry of the balanced chemical equation, we can calculate the theoretical yield of ammonia that can be produced from the given amount of nitrogen gas:

Moles of nitrogen gas = 0.5 mol

Moles of ammonia gas = 2 x moles of nitrogen gas = 2 x 0.5 mol = 1 mol

Mass of ammonia gas = Moles x molar mass = 1 mol x 17g/mol = 17g

Therefore, the maximum mass of ammonia that can be produced in the reaction is 17g.

Conclusion:

The maximum mass of ammonia that can be produced when 14g of nitrogen gas is mixed with 6g of hydrogen gas is 17g.