Calculate the change in entropy when 10.0 g of ice at -10.0°C is converted to water vapor at 115 °C at a constant pressure of 1 bar,.
=
Information: Cp, m(H20, 3) = +37.6 J K-1 mol-1; Cp, m(H20, 2) = + 75.75.3 J K-1 mol-1; Cp, m(H20,
g) = +33.6 J K mol1. The standard enthalpy of vaporization of H20 (1) is +40.7 kJ mol-1
and the standard enthalpy of fusion of H2O(s) = 6.01 kJ mol-1. At 1 bar Tmp = 0.0 °C and Tbp = 100.0°C.

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

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since electrons are negative the answer is neagtaive