Hydrogen can react with nitrogen to form ammonia. Which molecule has the largest

Question

Hydrogen can react with nitrogen to form ammonia. Which molecule has the largest numerical coefficient in the balanced equation? Hydrogen Nitrogen Ammonia more than one has the largest # Hydrogen can react with nitrogen to form ammonia. If I react 11.0 grams of hydrogen with 75.0 grams of nitrogen, how many grams of NH_3 with be made? 91.2 g 62.0 g 153 g 29.2 g Hydrogen can react with nitrogen to form ammonia. If I react 11.0 grams of with 75.0 grams of nitrogen, how many grams of excess chemical are used? 11.0 g 51.0 g 24.0g none of the above Hydrogen can react with nitrogen to form ammonia. If I react 11.0 grams of with 75.0 grams of nitrogen, how many grams of excess chemical are left over? 11.0 g 51.0 g 24.0 g none of the above What are the bond angle (s) in Xenon tetrachloride? 90, 120 degree 180 degree 90 degree 120 degree What is the hybridization of Xenon (Xe) in Xenon tetrachloride? sp^2 sp^3 dsp^3 d^2 sp^3

Explanation / Answer

1. N2 + 3H2 >>> 2NH3

(a)hydrogen is the answer 3 mols. highest coefficenit

2. Determine the limiting reagent:

N2 75.0 g / 28 g/mol = 2.67 mol
H2 11.0 g / 2 g/mol = 5.5 mol

N2 2.67 mol / 1 =2.67
H2 5.5 mol / 3 = 1.83

Seems pretty obvious that HYDROGEN is the limiting reagent.

2) Use H2 : NH3 molar ratio:

3 is to 2 as 5.500 mol is to x

x = 3.66 mol of H3 produced

3) Convert to grams:

3.66 mol x 17 g/mol = ~62 g g (to three sig fig)

3. Determine moles, then grams of NITROGEN used to hydrogen:

1 is to 3 as x is to 5.5 mol

x = 1.83 mol

1.83 mol times 28 g/mol of N2 = 51.24 g (b)

3) Determine grams of nitrogen remaining:

75 g g - 51.24 g = 23.86 g

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