Question

To convert from a given quantity of one reactant or product to the quantity of another reactant or product: First, convert the given quantity to moles. Use molar masses to convert masses to moles, and use Avogadro's number (6.02×1023 particles per mole) to convert number of particles to moles. Next, convert moles of the given reactant or product to moles of the desired reactant or product using the coefficients of the balanced chemical equation. For example, in the chemical equation 2H2+O2→2H2O the coefficients tell us that 2 mol of H2 reacts with 1 mol of O2 to produce 2 mol of H2O. Finally, convert moles of the desired reactant or product back to the desired units. Again, use molar masses to convert from moles to masses, and use Avogadro's number to convert from moles to number of particles.

Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation 3H2(g)+N2(g)→2NH3(g)

NOTE: Throughout this tutorial use molar masses expressed to five significant figures.

Part A: How many moles of NH3 can be produced from 21.0 mol of H2 and excess N2?

Part B: How many grams of NH3 can be produced from 4.99 mol of N2 and excess H2.

Part C: How many grams of H2 are needed to produce 12.43 g of NH3?

Part D: How many molecules (not moles) of NH3 are produced from 4.35×10−4 g of H2?

Answer 1

Part-A- from the balanced chemical equation it is clear that 3 moles of H2 produces 2 moles of NH3. Therefore we can consider

3 moles of H2 = 2 moles of NH3

For 21 moles of H2 = (21 x 2) / 3 = 14 moles of NH3

therefore, 14 moles of NH3 will be produced

Part-B- let us first calculate the number of moles of NH3 can be produced from 4.99 moles of N2.

From the balanced equation, 1 moles of N2 produces 2 moles of NH3. Therefore we can consider

1 moles of N2 = 2 moles of NH3

For 4.99 moles of N2 = 4.99 x 2 = 9.98 moles NH3

The molar mass of NH3 is 17.031 g/mol. The mass can be calculated using following formula

Mass of NH3 = number of moles of NH3 x molar mass of NH3

= 9.98 mol x 17.031 g/mol

= 169.97 g

Therefore, 169.97 g of NH3 will be produced.

Part-C- Let us first calculate the number of moles of NH3 in 12.43 grams. The molar mass of NH3 is 17.031 g/mol

Number of moles of NH3 = Weight / molar mass

= 12.43 g / 17.031 g/mol

= 0.73 moles of NH3

Now as we know,

2 moles of NH3 = 3 moles of H2

For   0.73 moles of NH3 =( 0.73 x 3) / 2 = 1.094 moles of H2

The molar mass of H2 is 2.01588 g/mol

mass of H2 = number of moles of H2 x molar mass of H2

= 1.094 mol x 2.01588 g/mol

= 2.2 g of H2

Therefore, 2.2 g of H2 will be needed.

Part-D- Let us first calculate the number of moles of H2 in 4.35 x 10-4 g = 0.000435.

Number of moles of H2 = weight / Molar mass

= 0.000435 g / 2.01588 g/mol

= 0.00021578 moles

as we know that,

3 moles of H2 = 2 moles of NH3

0.00021578 moles of H2 = (0.00021578 x 2) / 3 moles of NH3

= 0.00014385 moles of NH3

Now the number of molecules of NH3 can be calculated by multiplying number of moles of NH3 with avagadro's number

number of molecules of NH3 = 0.00014385 x 6.02 x 10²³

= 8.66 x 10¹⁹ molecules

Therefore, the answer is 8.66 x 10¹⁹ molecules of NH3