What coefficient should go in front of LiOH when the reaction is correctly balanced? Fe(OH)3 + Li2S --> LiOH + Fe2S3

6

Explanation:

Balancing a chemical equation involves making sure that the number of atoms of each element on the reactant side is equal to that on the product side. For example, if there are 5 moles of carbon on the reactant side, there must be 5 moles of carbon on the product side. This is in accordance to the law of conservation of mass.

However, to do the above, we make use of coefficients i.e whole numbers placed behind the elements. According to this question, an unbalanced chemical reaction is given as follows:

Fe(OH)3 + Li2S → LiOH + Fe2S3

As observed, the number of atoms on each side of the reaction are unequal. Hence, coefficients are used to balance the equation as follows:

2Fe(OH)3 + 3Li2S → 6LiOH + Fe2S3

From the balanced equation, 6 is used as a coefficient in front of LiOH.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol