Solution :
From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .
Moles of KBr in 102 g of potassium bromide.
n = 102/119.002
n = 0.86 mole.
So, number of miles of KCl produced are also 0.86 mole.
Mass of KCl produced :

Hence, this is the required solution.
Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
<u>Answer:</u> The reaction proceeds in the forward direction
<u>Explanation:</u>
For the given chemical equation:

Relation of
is given by the formula:

where,
= equilibrium constant in terms of partial pressure = ?
= equilibrium constant in terms of concentration = 
R = Gas constant = 
T = temperature = ![35^oC=[35+273]K=308K](https://tex.z-dn.net/?f=35%5EoC%3D%5B35%2B273%5DK%3D308K)
= change in number of moles of gas particles = 
Putting values in above equation, we get:

is the constant of a certain reaction at equilibrium while
is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
The expression of
for above equation follows:

We are given:



Putting values in above equation, we get:

We are given:

There are 3 conditions:
- When
; the reaction is product favored. - When
; the reaction is reactant favored. - When
; the reaction is in equilibrium
As,
, the reaction will be favoring product side.
Hence, the reaction proceeds in the forward direction
The number of moles of NH3 that could be made would be 0.5 moles
<h3>Stoichiometric reactions</h3>
From the balanced equation of the reaction:
N2 (g) + 3 H2(g) ----> 2NH3 (g)
The mole ratio of N2 to H2 is 1:3
Thus, for 0.50 moles of N2, 1.5 moles of H2 should be present. But 0.75 moles of H2 was allowed to react. Meaning that H2 is limiting in this case.
Mole ratio of H2 and NH3 = 3:2
Thus for 0.75 moles H2, the mole of NH3 that would be produced will be:
2 x 0.75/3 = 0.5 moles
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer:
An oxyacetylene torch can also be used for welding. When welding with an oxyacetylene torch, the flame is used to produce molten metal along the edge of two work pieces.
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