Milk of magnesia, Mg(OH)2, is prepared from the reaction of aqueous magnesium sulfate and sodium hydroxide solution. If a solution containing 125 g of MgSO4 is added to a solution with 115 g of NaOH, what is the mass of milk of magnesia produced

Explanation:

Hello.

In this case, since the described chemical reaction between aqueous magnesium sulfate and sodium hydroxide is:

Whereas can see a 2:1 mole ratio between sodium hydroxide and magnesium sulfate whose molar masses are 120.366 g/mol and 39.997 g/mol. Next step, is to compute the yielded moles of milk of magnesia yielded by both 125 g of magnesium sulfate (1:1 mole ratio) and 115 g of sodium hydroxide (2:1 mole ratio):

Thus, since magnesium sulfate yields less moles of milk of magnesia than sodium hydroxide, we infer that MgSO4 is the limiting reactant so the yielded mass of milk of magnesia (58.320 g/mol) is:

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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

the temperature of the gas, in k, not celsius.