Sulfur trioxide, ammonia gas, and water will react to form ammonium sulfate. Determine the percent yield when 3.78 g of ammonium sulfate forms from the reaction of 2.51 g of sulfur
trioxide, 1.55 g of ammonia, and 0.804 g of water.

Explanation:

Sulfur trioxide, ammonia gas, and water will react to form ammonium sulfate. Determine the percent yield when 3.78 g of ammonium sulfate forms from the reaction of 2.51 g of sulfur trioxide, 1.55 g of ammonia, and 0.804 g of water. Can you please show step by step on how you got the answer thanks!

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First, the balanced equation:

1 SO3 + 2 NH3 + 1 H2O --> 1 (NH4)2SO4

Find the limiting reactant using molar masses and mole ratios

2.51 g SO3 (1 mol SO3 / 80.06 g SO3) (1 mol (NH4)2SO4 / 1 mol SO3) = 0.03135 mol (NH4)2SO4

1.55 g NH3 (1 mol NH3 / 17.04 g NH3) (1 mol (NH4)2SO4 / 2 mol NH3) = 0.04548 mol (NH4)2SO4

0.804 g H2O (1 mol H2O / 18.02 g H2O) (1 mol (NH4)2SO4 / 1 mol H2O) = 0.04462 mol (NH4)2SO4

Therefore, the limiting reactant is sulfur trioxide.

Now find the theoretical yield of (NH4)2SO4.

2.51 g SO3 (1 mol SO3 / 80.06 g SO3) (1 mol (NH4)2SO4 / 1 mol SO3) (132.16 g (NH4)2SO3 / 1 mol (NH4)2SO4) = 4.14 g (NH4)2SO4

Actual yield / Theoretical yield x 100% = Percent yield.

Therefore, percent yield = 91.3%

hoppe it helps

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.