Answer:- 14.0 moles of hydrogen present in 2.00 moles of
.
Solution:- We have been given with 2.00 moles of
and asked to calculate the grams of hydrogen present in it. It's a two step conversion problem. In first step we convert the moles of the compound to moles of hydrogen as one mol of the compound contains 7 moles of hydrogen. In next step the moles are converted to grams on multiplying the moles by atomic mass of H. The calculations are shown as:

= 14.0 g H
So, there are 14.0 g of hydrogen in 2.00 moles of
.
Based on Le Chatelier's principle, if a system at equilibrium is disturbed by changes in the temperature, pressure or concentration, then the equilibrium will shift in a direction to undo the effect of the induced change.
The given reaction is endothermic i.e, heat is supplied:
CH4(g) + H2O (g) + heat ↔ 3H2(g) + CO(g)
a) When the temperature is lowered, heat is being removed from the system. The reaction will move in a direction to produce more heat i.e. to the left.
Hence, the pressure of CH4 will increase and equilibrium will shift to the left
b) When the temperature is raised, heat is being added to the system. The reaction will move in a direction to consume the added heat i.e. to the right.
Hence, the pressure of CO will increase and equilibrium will shift to the right
Answer:
7.82 g of Cu
Explanation:
2 moles of Al react to 3 moles of copper sulfate in order to produce 3 moles of copper and 1 mol of aluminum sulfate.
Firstly we determine the moles of reactant.
As copper sulfate is in excess, Al is the limiting.
2.75 g . 1mol /26.98g = 0.102 moles
Ratio is 2:3. 2 moles of Al, can produce 3 moles of Cu
So the 0.102 moles of Al will produce(0.102 . 3) /2 = 0.153 moles.
We convert moles to mass: 0.153 mol . 63.5g /mol = 9.71 g
That's the theoretical yield (100 % yield reaction)
We know that: (yield produced / theoretical yield) . 100 = percent yield
We replace:
(Yield produced / 9.71g) . 100 = 80.5 %
(Yield produced / 9.71g) = 0.805
Yield produced = 0.805 . 9.71g = 7.82 g