Question #180183

Use the following balanced equation to answer the parts below:

3H2 + N2 2NH3

How many grams of NH_{3} can be made from 5.60 mol of H_{2} and excess N_{2}?

How many grams of NH_{3} can be made from 4.34 mol of N_{2} and excess H_{2}?

If you want to make 19.59 mol of NH_{3}, how many grams of N_{2} do you need?

Expert's answer

Molar Mass of NH_{3 }= 17.03

Molar Mass of N_{2} = 28.0134

Molar Mass of H_{2} = 2.016

How many grams of NH_{3} can be made from 5.60 mol of H_{2} and excess N_{2}?

From the equation

= 3/4×5.6 = 4.2moles of H_{2}

= 1/4 × 5.6 = 1.4 moles of N_{2}

3moles of H_{2 }= 2 moles of NH_{3}

2/3 × 4.2 = 2.8 moles of NH_{3}

= 2.8 × 17.031 = 47.6868g

1 mole of N_{2 }= 2 moles of NH_{3}

= 1.4 × 2 = 2.8 × 17.031= 47.6868g

= 47.6868 + 47.6868 = 95.3736g of NH_{3}

How many grams of NH_{3} can be made from 4.34 mol of N_{2} and excess H_{2}?

= 3/4×4.34 = 3.255 moles

1/4×4.34 = 1.085moles

= 2/3×3.255 = 2.17 moles of NH_{3 }

= 2.17×17.031 = 36.95727g

1.085 × 2 = 2.17

= 36.95727g

= 36.95727×2 = 73.9145g of NH_{3}

If you want to make 19.59 mol of NH_{3}, how many grams of N_{2} do you need?

19.59 moles of NH_{3}

2 moles of NH_{3 }= 1mole of N_{2}

= 19.59/2 = 9.795× 28.0134

= 274.391253g of N_{2}

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