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4NH3 +5O2 → 4NO +6H2O

Using the equation above,what is the limiting reagent and theoretical yield in grams of nitric oxide when using 4.65g of ammonia to react with 4.25g of oxygen? Also How many grams of excess reagent remain? Show all of the steps and show all of them step by step, and also show SIG FIGS if necessary.

4. Silicon reacts with nitrogen to form silicon nitride (SiN4).
Write and show all of the necessary work and all of the steps for a balanced equation, and also show SIG FIGS if necessary. Also show How many grams of silicon are needed to form 98.5 grams of silicon nitride? Identify the limiting and excess reagents when there are 40.0g of silicon and 25.5g of nitrogen. How many additional grams of nitrogen do we need to produce 98.5 grams of product? Show all of the steps and show all of the work.

5. Solid sulfur and oxygen gas react to form sulfur trioxide gas. Write and show work for the balanced chemical equation with correct physical states. Show all of the steps!! For both of the problems! If 150.2g of oxygen reacted with 105.0g of sulfur and the actual yield in the lab was 246.6g of sulfur trioxide. What is the theoretical yield in grams? Show all of the steps!

User Icon for Bot GPT 3.5 Bot GPT 3.5 answered
4 months ago

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.