The mass of sodium hydroxide produced when 175.5 g
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Na(s)+H2O(l)→NaOH(aq)+12H2(g)
Moles of NaOH = 20.0⋅g22.99⋅g⋅mol−1 = ?? mol
Explanation:
Clearly, there is a 1:1 equivalence between the moles of sodium, and moles of sodium hydroxide. One half an equiv of dihydrogen gas is also produced. We know that 1 mol of gas occupies 24.5 L at 298 K, and 1 atm. What volume of dihydrogen would be evolved under the above conditions?
Moles of NaOH = 20.0⋅g22.99⋅g⋅mol−1 = ?? mol
Explanation:
Clearly, there is a 1:1 equivalence between the moles of sodium, and moles of sodium hydroxide. One half an equiv of dihydrogen gas is also produced. We know that 1 mol of gas occupies 24.5 L at 298 K, and 1 atm. What volume of dihydrogen would be evolved under the above conditions?
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