3H2 + N2 >> 2NH3 A. How many miles of NH3 can be produced from 16.5 mol H2 and e
ID: 968481 • Letter: 3
Question
3H2 + N2 >> 2NH3 A. How many miles of NH3 can be produced from 16.5 mol H2 and excess N2? B. How many grams of NH3 can be produced from 2.30 mol N2 and excess H2? C. How many grams of H2 are needed to produce 12.20g NH3? D. How many molecules (not moles) of NH3 are produced from 8.32x10^-4g H2? 3H2 + N2 >> 2NH3 A. How many miles of NH3 can be produced from 16.5 mol H2 and excess N2? B. How many grams of NH3 can be produced from 2.30 mol N2 and excess H2? C. How many grams of H2 are needed to produce 12.20g NH3? D. How many molecules (not moles) of NH3 are produced from 8.32x10^-4g H2? A. How many miles of NH3 can be produced from 16.5 mol H2 and excess N2? B. How many grams of NH3 can be produced from 2.30 mol N2 and excess H2? C. How many grams of H2 are needed to produce 12.20g NH3? D. How many molecules (not moles) of NH3 are produced from 8.32x10^-4g H2?Explanation / Answer
A. We use stoichiometric relation:
16.5 mol H2 * (2 mol NH3 / 3 mol H2) = 11 moles of NH3
B. First we use stoichiometric relation:
2.30 mol N2 * (2 mol NH3 / 1 mol N2) = 4.6 mol NH3
Then we use molar mass:
4.6 mol NH3 * (17g / mol) = 78.2 grams of NH3
C. First we turn grams to moles of ammonia:
12.20 g of NH3 * (1mol / 17g) = 0.7176 moles
Then we use stoichiometric relation:
0.7176 moles of NH3 * (3 moles of H2 / 2 moles NH3) = 1.0765 moles of H2
1.0765 mol H2 * (2 grams / mol) = 2.1529 grams of H2
D. First we pass grams to moles:
8.32 x 10-4 g of H2 * (1 mol / 2 grams) = 4.16 x 10-4 moles of H2
We then use stoichiometric relation:
4.16 x 10-4 moles of H2 * (2 moles of NH3 / 3 moles of H2) = 2.7733 x 10-4 moles of NH3
Finally we pass them to molecules with avogadro's number:
2.7733 x 10-4 moles of NH3 * (6.022 x 1023 molecules / mol) = 1.67 x 1020 molecules
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