How many grams of hydrogen are necessary to react completely with 50.0 g of nitrogen in the reaction #N_2+3H_2->2NH_3#?

Let's compute the mole ratio using the equation.

One mole of nitrogen reacts with one mol of hydrogen, according to the equation.

In mass terms, 3 mol, or 6.048 g, of hydrogen are required for every 28.01 g of nitrogen.

We could configure the ratio;

Assume that: n = number of moles m = mass of substance M = molar mass (equivalent to atomic weight on the periodic table)

The next step is to find out the molar mass (M) of Hydrogen. If you refer to your periodic table, you know that hydrogen's molar mass is 1.0 g/mol.

Since there are two hydrogens present the molar mass of hydrogen in this reaction is 2.0 g/mol. (look at equation, carefully - there are 2, not 5 or 6 because the front number is only used for moles).

Your last step is to find out its mass (m).

Therefore, 10.7 grams of Hydrogen are necessary to fully react with 50.0 grams of Nitrogen.

To solve this problem, we first need to determine the molar masses of nitrogen (N₂) and hydrogen (H₂), which are 28.02 g/mol and 2.016 g/mol, respectively.

Next, we need to calculate the number of moles of nitrogen present in 50.0 g using the formula:

[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} ]

[ \text{moles of N₂} = \frac{50.0 , \text{g}}{28.02 , \text{g/mol}} ]

[ \text{moles of N₂} \approx 1.78 , \text{mol} ]

According to the balanced chemical equation, 1 mole of nitrogen reacts with 3 moles of hydrogen. Therefore, the number of moles of hydrogen required can be calculated as:

[ \text{moles of H₂} = 3 \times \text{moles of N₂} ]

[ \text{moles of H₂} = 3 \times 1.78 , \text{mol} ]

[ \text{moles of H₂} \approx 5.34 , \text{mol} ]

Finally, we can calculate the mass of hydrogen required using its molar mass:

[ \text{mass of H₂} = \text{moles} \times \text{molar mass} ]

[ \text{mass of H₂} = 5.34 , \text{mol} \times 2.016 , \text{g/mol} ]

[ \text{mass of H₂} \approx 10.77 , \text{g} ]

Therefore, approximately 10.77 grams of hydrogen are necessary to react completely with 50.0 grams of nitrogen.