Here is a site on Jiskha where I worked through the steps for a limiting reagent problem. Follow the steps.
http://www.jiskha.com/display.cgi?id=1270504808
5F2(g) + 2NH3(g) ==> N2F4(g) + 6HF(g)
a)If you have 66.6 g NH3, how many grams of F2 are required for complete reaction?
b)How many grams of NH3 are required to produce 4.65g HF?
c)How many grams of N2F4 can be produced from 225g F2?
http://www.jiskha.com/display.cgi?id=1270504808
a) To find the number of grams of F2 required for complete reaction with 66.6 g NH3, we need to use the molar ratio between F2 and NH3 in the balanced equation.
Step 1: Convert grams of NH3 to moles of NH3.
Given: 66.6 g NH3
Molar mass of NH3 = 17.03 g/mol (N*1 + H*3)
Moles of NH3 = (66.6 g NH3) / (17.03 g/mol NH3)
Step 2: Use the molar ratio between F2 and NH3 from the balanced equation.
Given: 5 moles of F2 react with 2 moles of NH3
Moles of F2 = [Moles of NH3 * (5 moles F2 / 2 moles NH3)]
Step 3: Convert moles of F2 to grams of F2.
Given: Molar mass of F2 = 38.0 g/mol
Grams of F2 = (Moles of F2) x (Molar mass of F2)
b) To find the number of grams of NH3 required to produce 4.65 g of HF, we need to use the molar ratio between NH3 and HF.
Step 1: Convert grams of HF to moles of HF.
Given: 4.65 g HF
Molar mass of HF = 20.01 g/mol (H*1 + F*1)
Moles of HF = (4.65 g HF) / (20.01 g/mol HF)
Step 2: Use the molar ratio between NH3 and HF from the balanced equation.
Given: 2 moles of NH3 produce 6 moles of HF
Moles of NH3 = [Moles of HF * (2 moles NH3 / 6 moles HF)]
Step 3: Convert moles of NH3 to grams of NH3.
Given: Molar mass of NH3 = 17.03 g/mol
Grams of NH3 = (Moles of NH3) x (Molar mass of NH3)
c) To find the number of grams of N2F4 that can be produced from 225 g of F2, we need to use the molar ratio between F2 and N2F4.
Step 1: Convert grams of F2 to moles of F2.
Given: 225 g F2
Molar mass of F2 = 38.0 g/mol
Moles of F2 = (225 g F2) / (38.0 g/mol F2)
Step 2: Use the molar ratio between F2 and N2F4 from the balanced equation.
Given: 5 moles of F2 produce 1 mole of N2F4
Moles of N2F4 = [Moles of F2 * (1 mole N2F4 / 5 moles F2)]
Step 3: Convert moles of N2F4 to grams of N2F4.
Given: Molar mass of N2F4 = 104.0 g/mol
Grams of N2F4 = (Moles of N2F4) x (Molar mass of N2F4)
Using these calculations, you can find the answers to the given questions by substituting the appropriate values.