In the reaction, 2Hg(l) + O2(g) ---> 2HgO(s), calculate the amount, in grams, of HgO formed when 20.0 g of Hg react with 25.0g O2
please explain the steps as well. Thanks
Dr.A
Favorite Answer
moles O2 = 25.0 g/ 32 g/mol=0.781
the ratio between O2 and Hg is 1 : 2
moles Hg required = 0.781 x 2 =1.56
but we have only 20.0 g/ 200.59 g/mol=0.0997 moles of Mercury so it is the limiting reactant
the ratio between Hg and HgO is 2 : 2
moles HgO = 0.0997
mass HgO = 0.0997 mol x 216.59 g/mol=21.6 g
Anonymous
Al2S3 + 6H2O --> 2Al(OH)3 + 3H2S Having the balanced equation describing the reaction, the subsequent step is calculating the form of moles. The molecular weight of Al2S3 is what one million mole of its molecules weighs. this may be the sum of the atomic weights of the atoms interior the molecule. For Al2S3, the load may be 27*2 + 32*3 = a hundred and fifty grams in step with mole. as a result 200 grams may be 200 grams / (a hundred and fifty grams/mole) = one million.333 moles. The form of moles of water obtainable is a hundred and fifty grams / (18 grams/mole) = 8.33 moles. Now, examine the molar ratios from the balanced equation. one million mole of Al2S3 demands 6 moles of H2O to variety 3 moles of H2S. From the calculations above, one million.333 moles Al2S3 might require 8 moles of water. There are 8.33 moles obtainable, so there is extra effective than adequate to variety the product. The molar ratio of Al2S3 to H2S is one million to 3, so one million.33 moles will variety 4 moles of H2S. 4 moles of H2S weighs 4 moles * (34 grams/ mole) = 136 grams.