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%.
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?
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