Answer:
34.44 g.
Explanation:
- The balanced equation for the reaction is:
<em>Fe + S → FeS,</em>
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<em>It is clear that 1.0 mol of Fe reacts with 1.0 mol of S to produce 1.0 mol FeS.</em>
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- We need to calculate the no. of moles of 60 grams of iron and 90 grams of iron sulfide:
no. of moles of Fe = mass/atomic mass = (60.0 g)/(55.845 g/mol) = <em>1.074 mol.</em>
<em>no. of moles of FeS = mass/molar mass</em> = (90.0 g)/(87.91 g/mol) = <em>1.024 mol.</em>
<u><em>∵ Fe reacts with S with (1: 1) molar ratio.</em></u>
∴ The no. of moles of S needed to react with Fe is 1.074 mol.
<em>∴ The no. of grams of S needed = no. of moles x molar mass </em>= (1.074 mol)(32.065 g/mol) = <em>34.44 g.</em>
When Q is equal the initial concentration of the products / the initial concentration of the reactants.
so, Q = [Ag]*[Cl-] and we neglected [AgCl] as it is solid
∴ Q = 10^-6 * 10^-5
= 10^-11
now we will compare the value of Q with the value of Keq:
when Q = Keq so, the system is in equilibrium
and when Q > Keq so, the reaction will go forward (shift to right) to achieve equilibrium.
and when Q< Keq so, the reaction will go left (shift to left) to achieve equilibrium.
when Q = 10^-11 and Keq = 10^20
∴Q< Keq
and the reaction will shift to left.
Answer:
A higher concentration of a catalyst will speed up the reaction rate.
