4NH3 + 5O2 → 4NO + 6H2O
Step 1: Convert the mass of each reactant to moles
4.65g NH3 x (1 mol NH3/17.03g NH3) = 0.273 moles NH3
4.25g O2 x (1 mol O2/32.00g O2) = 0.133 moles O2
Step 2: Determine the limiting reagent by calculating the mole ratio between NH3 and O2
From the balanced equation, the ratio of NH3 to O2 is 4:5
NH3: 0.273 moles
O2: 0.133 moles
Since the mole ratio is 4:5, O2 is the limiting reagent
Step 3: Calculate the theoretical yield of NO in grams
Using the mole ratio between O2 and NO (5:4)
0.133 moles O2 x (4 moles NO/5 moles O2) x (30.01g NO/1 mol NO) = 0.96g NO
Step 4: Calculate the grams of excess reagent (NH3) that remain
0.273 moles NH3 (17.03g NH3/1 mol NH3) - 4.65g NH3 = 0.36g NH3
Therefore, the limiting reagent is O2 with a theoretical yield of 0.96g NO, and 0.36g of NH3 remains as the excess reagent.
2Si + 2N2 → 2SiN4
Step 1: Convert mass of silicon and nitrogen to moles
40.0g Si x (1 mol Si/28.09g Si) = 1.42 moles Si
25.5g N2 x (1 mol N2/28.02g N2) = 0.91 moles N2
Step 2: Determine the limiting reagent by calculating the mole ratio between Si and N2
From the balanced equation, the ratio of Si to N2 is 2:2 (1:1)
Si: 1.42 moles
N2: 0.91 moles
Since the mole ratio is 1:1, both reactants are in the stoichiometric ratio.
Step 3: Calculate the theoretical yield of SiN4 in grams
Using the mole ratio between Si and SiN4 (1:1)
1.42 moles Si x (1 mol SiN4/1 mol Si) x (104.1g SiN4/1 mol SiN4) = 148g SiN4
Step 4: Calculate additional grams of N2 needed to produce 98.5g of SiN4
98.5g SiN4 x (1 mol SiN4/104.1g SiN4) x (1 mol N2/2 mol SiN4) x (28.02g N2/1 mol N2) = 14.8g N2
Therefore, no additional grams of N2 are needed to produce 98.5g of SiN4.
2S + 3O2 → 2SO3
Step 1: Convert the mass of sulfur to moles
105.0g S x (1 mol S/32.06g S) = 3.27 moles S
Step 2: Determine the limiting reagent by calculating the mole ratio between S and O2
From the balanced equation, the ratio of S to O2 is 2:3
S: 3.27 moles
O2: 150.2g O2 x (1 mol O2/32.00g O2) = 4.70 moles O2
Since the mole ratio is 2:3, S is the limiting reagent.
Step 3: Calculate the theoretical yield of SO3 in grams
Using the mole ratio between S and SO3 (2:2)
3.27 moles S x (2 mol SO3/2 mol S) x (80.06g SO3/1 mol SO3) = 261.6g SO3
Therefore, the theoretical yield of sulfur trioxide is 261.6g.