According to the equation H2SO4+2NH3 @(NH4)2SO4. What mass of ammonium sulfate will result from the reaction of 5.00 g of NH3?
no
To find the mass of ammonium sulfate produced from the reaction of 5.00 g of NH3, we need to use the balanced equation and calculate the molar ratio.
The balanced equation for the reaction is:
H2SO4 + 2NH3 → (NH4)2SO4
First, we need to find the number of moles of NH3 given the mass provided.
Step 1: Calculate the molar mass of NH3.
The molar mass of NH3 is:
1(1.01 g/mol) + 3(1.01 g/mol) = 17.03 g/mol
Step 2: Calculate the number of moles of NH3.
Using the given mass of NH3 (5.00 g) and the molar mass of NH3 (17.03 g/mol):
Number of moles = Mass / Molar mass
Number of moles = 5.00 g / 17.03 g/mol
Number of moles ≈ 0.293 mol
Step 3: Calculate the molar ratio between NH3 and (NH4)2SO4.
From the balanced equation, the ratio between NH3 and (NH4)2SO4 is 2:1. This means that 2 moles of NH3 react to produce 1 mole of (NH4)2SO4.
Step 4: Calculate the number of moles of (NH4)2SO4 produced.
Given the moles of NH3 (0.293 mol) and the molar ratio of 2:1, we can calculate the moles of (NH4)2SO4 produced:
Number of moles of (NH4)2SO4 = Moles of NH3 / Molar ratio
Number of moles of (NH4)2SO4 = 0.293 mol / 2
Number of moles of (NH4)2SO4 ≈ 0.146 mol
Step 5: Calculate the mass of (NH4)2SO4 produced.
Using the calculated moles of (NH4)2SO4 (0.146 mol) and the molar mass of (NH4)2SO4:
Molar mass of (NH4)2SO4 = 2(14.01 g/mol) + 4(1.01 g/mol) + 32.06 g/mol = 132.14 g/mol
Mass of (NH4)2SO4 = Number of moles × Molar mass
Mass of (NH4)2SO4 = 0.146 mol × 132.14 g/mol
Mass of (NH4)2SO4 ≈ 19.25 g
Therefore, the mass of ammonium sulfate resulting from the reaction of 5.00 g of NH3 is approximately 19.25 g.
To find the mass of ammonium sulfate (NH4)2SO4 produced from the reaction of 5.00 g of NH3, we need to use stoichiometry and molar ratios.
1. Write down the balanced chemical equation:
H2SO4 + 2NH3 → (NH4)2SO4
2. Determine the molar mass of NH3:
The molar mass of NH3 is 17.03 g/mol.
3. Convert the given mass of NH3 to moles:
Moles = Mass / Molar mass
Moles of NH3 = 5.00 g / 17.03 g/mol
4. Use the balanced equation to establish the mole ratio between NH3 and (NH4)2SO4:
From the equation, 2 moles of NH3 produce 1 mole of (NH4)2SO4.
5. Calculate the moles of (NH4)2SO4 produced:
Moles of (NH4)2SO4 = Moles of NH3 / 2
6. Use the molar mass of (NH4)2SO4 to convert moles to grams:
Mass = Moles x Molar mass
Mass of (NH4)2SO4 = Moles of (NH4)2SO4 x Molar mass of (NH4)2SO4
By following these steps, you can calculate the mass of ammonium sulfate resulting from the given amount of NH3.
ok..so write the balance equation;
H2SO4 + 2NH3 --> (NH4)2SO4
this shows that 1 mole of sulphuric acid react with 2 moles of NH3 to produce 1mole of ammonium sulphate. The question asks the mass of (NH4)2SO4 produced by 5.00g of NH3 is used. So focus only on these two; ignore sulphuric acid.
so, mole ratio
2:1 (2mole NH3 produces 1mole (NH4)2SO4).
here's what to do;
1. calculate the mole of NH3 using n = m/Mr where m is the mass in grams and Mr is the molar mass.
2. Use the mole of NH3 to determine the mole of ammonium sulphate using the mole ratio of the two.
3. Use the mole for ammonium sulphate to calculate the mass by using m = nMr where n is the mole you calculated in step 2 and Mr is the molar mass of ammonium sulphate.
Hope that helps.