Given the following reaction: Na2S2O3 + AgBr (yields) NaBr + Na3 [Ag(S2O3)2 ]
a. How many moles of Na2S2O3 are needed to react completely with 42.7 g of
AgBr?
b. What is the mass of NaBr that will be produced from 42.7 g of AgBr?

I need you to show the work please.

Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2]

Explanation:

First check that the given equation is balanced ... it isn't ... so the first thing to do is balance the equation:

balancing Na:

2Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2]

and everything is now balanced so we've got the balanced equation

molar mass AgBr = 107.87 + 79.90 = 187.77 g/mol

mol AgBr available = 42.7 g AgBr x [1 mol / 187.77 g] = 0.2274 mol AgBr

from the balanced equation the mole ratio AgBr : Na2S2O3 = 1 : 2

so mol Na2S2O3 required = 0.2274 mol AgBr x [ 2 mol Na2S2O3 / mol AgBr]

= 0.455 mol Na2S2O3 (to 3 sig figs)

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j