Answer:
4.37 g of barium sulphate
Explanation:
The reaction equation is;
3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)
From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles
To find the limiting reactant;
3 moles of barium chloride yields 3 moles of barium sulphate
0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate
1 mole of iron III sulphate yields 3 moles of barium sulphate
0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate
Hence,barium chloride is the limiting reactant
Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate
Answer:
For large rivers the problem is not simply a matter of deduction of consumptive use from runoff: it is more complex and the complexity is related to the changes in .
Explanation:
There is an excess of oxygen, the two main products should be carbon dioxide gas and water vapor. If the oxygen supply is limited, the fuel will undergo incomplete combustion and produce carbon monoxide, water, and sometimes carbon.
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