14.0g of F2 was reacted with excess NH3 to produce N2F4 and HF. The equation for the reaction: 5F2(g) + 2NH3(g) - N2F4(g) + 6HF(g)
a.) How many grams of N2F4 can theoretically be present?
b.) If 6.82g of N2F4 is obtained from the experiment what is the percent yield?
To find the answer, we need to use stoichiometry and the information given in the problem.
a.) According to the balanced equation, 5 moles of F2 reacts with 2 moles of NH3 to produce 1 mole of N2F4. This means that the molar ratio between F2 and N2F4 is 5:1.
Therefore, we can calculate the theoretical yield of N2F4 by converting the mass of F2 to moles, using its molar mass (38.0 g/mol), and then using the molar ratio:
14.0 g F2 * (1 mol F2 / 38.0 g F2) * (1 mol N2F4 / 5 mol F2) = 0.0737 mol N2F4
To find the mass of N2F4, we multiply the number of moles by its molar mass:
0.0737 mol N2F4 * 92.0 g/mol = 6.78 g N2F4
Therefore, the maximum theoretical yield of N2F4 is 6.78 g.
b.) The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100:
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (6.82 g / 6.78 g) * 100 = 100.6%
Therefore, the percent yield is 100.6%.