Enter your answer in the provided box. To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night

409.0 kg of sodium sulfate decahydrate will produce 4.49×10⁵ kJ

of heat energy.

Explanation:

CHECK THE COMPLETE QUESTION BELOW

To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night? Note that sodium sulfate decahydrate will transfer 354 kJ/mol

EXPLANATION

Here we were asked to calculate the amount of heat will be generated by 409.0 kg of sodium sulfate decahydrate at night assuming there Isa complete reaction and 100% efficiency of heat transfer in the process

The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S) is needed here, so it must be firstly calculated.

The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S)

( 1*20) + (22.98*2) + (16*14)+ (32*14)= 322.186 g/mol.

Thus 409.0 kg of H₂₀Na₂O₁₄S will have a value which is equivalent to = (409000g)/(322.186 g/mol.)

=1269.453mol of H₂₀Na₂O₁₄S.

But it was stated in the the question that per mole of H₂₀Na₂O₁₄S will transfer 354 kJ heat.

Therefore, 1269.453mol will transfer 1269.453× 354 kJ = 4.49×10⁵ kJ of heat.

Hence, 409.0 kg of sodium sulfate decahydrate will produce

4.49×10⁵ kJ of heat energy.

air quality

poor drinking water

and heavy polution

can lead to poor health

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g