CH.04HW Question 2 previous1 2 Question 2 Converting between quantities To conve

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CH.04HW Question 2 previous1 2 Question 2 Converting between quantities To convert from a given quantity of one reactant or product to the quantity of another reactant or product Learning Goal: To understand how to use stoichiometry to convert between quantities of reactants and products in chemical First, convest the given quantity to moles. Use molar masses to convert masses to moles, and use Avogadio's number (6.02 x 10" paticles 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 uning the coefficients of the balanced Next, corwert roles of the given reactant or product to moles of the desired mactant or product useg the coeficients of the baseed chemical equation For example, in the chemical equation · the quanttalive reanced reaction among the reactants and products of a balanced reaction by drectly comparing mole ratios Stoichiometry can be used to convert mass, number of moles, or number of particles between products and reactants, as shown in the the coefficients tell us that 2 mol of Ha reacts with 1 mol of 0, to produce 2 mol of H 0 Finally convert moles of the desired reactant or product back to the desired units Again, use molar masses to convet from moles to masses, and use Avogadro's number to convert from moles to number of particles Reaction of hydrogen and nitrogen to form ammonia Hydogen gas, H, reacts with nitrogen gas. Na to form ammonia gas, NiH,, according to the equation 3112 (r) + N, (r)2NH,( ) NOTE: Throughout this tutorial use molar masses expressed to five significant figures Subslance A molar mas moles Part A How many moles of NH, can be produced frm 180 mol of H and excess N, Expeess your answer numerically in moles melar Subslance parnc @ e &|9

Explanation / Answer

part A )

N2 + 3H2 ---------------> 2NH3

           3                           2

           18                       ?

moles of NH3 = 18 x 2 / 3

                  = 12 moles

part B)

1 mol N2 gives -------------------> 34 g NH3

3.44 mol N2 gives --------------> 3.44 x 34

= 116.96 g

part C)

6 g H2 -------------------> 34 g NH3

x g H2 ------------------>12.73 g NH3

x = 6 x 12.73 /34

x = 2.25g

H2 mass = 2.25 g

6 g -----------------------> 2 mol NH3

7.89 x 10^-4 g H2 -----------> 7.89 x 10^-4 x 2 / 6 = 2.62 x 10^-4 moles

molecules = 6.023 x 10^23 x 2.62 x 10^-4

                 = 1.58 x 10^20

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