Answer:
0.912 mL
Explanation:
3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
FeCl3 is the limiting reactant.
Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles
Hence actual yield of Iron III sulphide = 0.043 moles
Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield
Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide
From the reaction equation,
2moles of iron III chloride produced 1 mole of iron III sulphide
x moles of iron III chloride, will produce 0.057 of iron III sulphide
x= 2× 0.057= 0.114 moles of iron III chloride
But
Volume= number of moles/ concentration
Volume= 0.114/0.125
Volume= 0.912 mL
Answer:
100 C or 212 F
Explanation:
That is the boiling point for liquid so both gas and liquid are presented.
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Assuming that 2H2O(l) is the product, the reactants would be 2H2(g) and O2(g). The number of moles in the reactants would be 2 moles of H2 and 1 mole of O2.
Answer:

Explanation:
Hello,
Based on the given reaction, since magnesium and water are in a 1:2 molar ratio at the reactants, we must apply the following stoichiometric factors to compute the complete reaction of the 6.0 g of magnesium:

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