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(NH₄)₂SO₄ = 14 * 2 + 1 * 8 + 32+ 16 * 4 => 132 amu
If you can a number of electrons in a atom you will get a ion of the element
1. The balanced equation tells us that 2 moles of H2S produce 2 moles of H2O.
8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced
8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically
% yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8 % yield
2. Calculate moles of each reactant.
150.0 g N2 x (1 mole N2 / 28.0 g N2) = 5.36 moles N2
32.1 g H2 x (1 mole H2 / 2.02 g H2) = 15.9 moles H2
The balanced equation tells us that we need 3 moles of H2 to react with every 1 mole of N2.
So if we have 5.36 moles N2, we need 3x that = 16.1 moles H2. Do we have that much available? No, just under at 15.9 moles. So H2 is the limiting reactant. At the end of the reaction there will be a little N2 left over.
Answer:
t=(70)ln(19.9/20)/-ln2
70 is the years. 19.9 kg is what is left after 0.1kg decay. 20kg is your initial amount.
t=0.506 years with a decay constant of 0.0099021/year
Explanation:
