How many grams of H2 are needed to produce 14.90 g of NH3?
How many molecules (not moles) of NH3 are produced from 1.59×10^(−4) g of H2?
How many molecules (not moles) of NH3 are produced from 1.59×10^(−4) g of H2?
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We need a balanced equation for the reaction: N2 + 3 H2 --->2 NH3. Note the 2:3 ratio of NH3:H2 according to the coefficients in the balanced equation.
14.90-g NH3 x (1 mole NH3 / 17.0-g NH3) x (3 mol H2 / 2 mol NH3) x (2.02-g H2/ 1 mol H2) =
2.66-g H2
1.59 x 10^-4-g H2 x ( 1 mol H2/ 2.02-g H2) x (2 mol NH3 / 3 mol H2) x 6.023 x 10^23 molec H2/ 1 mol H2) = 3.16 x 10^19 molecules of NH3
14.90-g NH3 x (1 mole NH3 / 17.0-g NH3) x (3 mol H2 / 2 mol NH3) x (2.02-g H2/ 1 mol H2) =
2.66-g H2
1.59 x 10^-4-g H2 x ( 1 mol H2/ 2.02-g H2) x (2 mol NH3 / 3 mol H2) x 6.023 x 10^23 molec H2/ 1 mol H2) = 3.16 x 10^19 molecules of NH3