9moIN2 :
The molar ratio between Nitrogen and ammonia is
1
:
3
, therefore, to produce 18 moles of ammonia, we will need
Answer:
Ammonia > Urea > Ammonium nitrate > Ammonium sulphate
Explanation:
Percentage by mass of nitrogen in NH3:
Molar mass of NH3= 17 g/mol
Hence % by mass = 14/17 × 100 = 82.35%
% by mass of NH4NO3
Molar mass of NH4NO3 = 80.043 g/mol
Hence; 28/80.043 × 100 = 34.98%
% by mass of (NH4)2SO4;
Molar mass of (NH4)2SO4= 132.14 g/mol
Hence; 28/132.14 × 100 = 21.19%
% by mass of CH4N2O
Molar mass of urea = 60.0553 g/mol
Hence 28/60.0553 × 100 = 46.62%
Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.
In this problem, we have density and we have mass so we can plug into the equation and solve for V.
38.6=270.2/V
<em>*Multiply both sides by V*</em>
38.6V=270.2
<em>*Divide both sides by 38.6*</em>
V=7
The volume of the gold nugget is 7cm3.
Hope this helps!!
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