if 68.0g of NH3 react with 304g of F2, how many grams of N2F4 are produced
This is a limiting reagent problem. I know that because amounts are given for BOTH reactants.
1. Write and balance the equation.
2a. Convert g NH3 to moles. moles = grams/molar mass.
2b. Same for g F2.
3a. Using the coefficients in the balanced equation, convert moles NH3 to moles N2F4
3b. Same procedure convert moles F2 to moles N2F4.
3c. Obviously, the numbers are different and both can't be right. The correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.
4. Using the smaller value for moles N2F4 convert moles to grams. g = moles x molar mass.
To find the amount of N2F4 produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product formed.
We can calculate the number of moles for each reactant using their molar masses:
Molar mass of NH3 (molecular weight) = 14.01 g/mol for N + 3(1.01 g/mol) for H = 17.03 g/mol
Number of moles of NH3 = 68.0 g / 17.03 g/mol = 3.99 moles
Molar mass of F2 = 2(19.00 g/mol) = 38.00 g/mol
Number of moles of F2 = 304 g / 38.00 g/mol = 8.00 moles
Next, we need to determine the stoichiometric ratio between NH3 and N2F4 using the balanced chemical equation:
4 NH3 + 3 F2 → 2 N2F4
The stoichiometry tells us that for every 4 moles of NH3, we produce 2 moles of N2F4. Therefore, using the mole ratio, we can calculate the theoretical yield of N2F4:
Moles of N2F4 = (moles of NH3 / 4) * 2 = (3.99 moles / 4) * 2 = 1.99 moles
Finally, we can calculate the mass of N2F4 using its molar mass:
Molar mass of N2F4 = 92.0 g/mol
Mass of N2F4 = moles of N2F4 * molar mass of N2F4 = 1.99 moles * 92.0 g/mol = 183.08 g
Therefore, approximately 183.08 grams of N2F4 are produced.