Aluminium reacts with dilute sulfuric acid based on the following reaction:
<span>2Al + 3H2SO4 ..............> Al2 (SO4)3 + 3H2
From the periodic table:
mass of aluminium = 27 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
mass of sulfur = 32 grams
Therefore:
molar mass of aluminium = 27 grams
molar mass of sulfuric acid = 2(1) + 32 + 4(16) = 98 grams
From the balanced chemical equation:
2 moles of aluminium react with 3 moles of dilute sulfuric acid.
This means that 34 grams of Al react with 294 grams of the acid
To get the amount of aluminium that reacts with </span><span>5.890 g of sulfuric acid, we will do cross multiplication as follows:
</span>amount of Al = (<span>5.890 x 34) / 294 = 0.6811 grams</span>
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%
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