化學問題[摩爾計算]

Question

1.要完全中和20cm^3的2.0M氨水,需用16.0cm^3硫酸。該硫酸濃度是多少??(以g dm^-3為單位)
2.某個MgSO4．xH2O(s)樣本質量為123.2g,它包含63.0g結晶水。x的值是多少?

Answer 1

1.
所用NH3的摩爾數 = MV = 2 x (20 x 10-3) = 0.04 mol

2NH3 + H2SO4 → (NH4)2SO4
摩爾比 NH3 : H2SO4 = 2 : 1
所用H2SO4的摩爾數 = 0.04 x (1/2) = 0.02 mol

H2SO4的摩爾質量 = 2x1 + 32.1 + 16x4 = 98.1 g mol-1
所用H2SO4的質量 = 摩爾數 x 摩爾質量 = 0.02 x 98.1 = 1.962 g
所用H2SO4的濃度 = 質量 / 體積 = 1.962 / (16 x 10-3) = 122.6 g dm-3

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2.
MgSO4的摩爾質量 = 24.3 + 32.1 + 16x4 = 120.4 g mol-1
H2O的摩爾質量 = 1x2 + 16 = 18 g mol-1
樣本中H2O的質量 = 63 g
樣本中MgSO4的質量 = 123.2 - 63 = 60.2 g

摩爾比 MgSO4 : H2O
= (60.2/120.4) : (63/18)
= 0.5 : 3.5
= 1 : 7

所以 x = 7

Answer 2

1)NH3+H2O <===>NH4OH
2(NH4OH)+H2SO4 ===>(NH4)2SO4+2(H2O)

2 mole ammonium water need 1 mole of H2SO4 to neutralize
no.moles of H2SO4 used = 0.5*no.moles of NH4OH used
=0.5*2*20/1000=0.02mol
mass of H2SO4 used=0.02/(1*2+32.1+16*4)=2.04*10^-4 g
concentration=2.04*10^-4/(16/1000)=0.0127421 g dm^-3



2)formula mass of MgSO4=24.3+32.1+16*4=120.4
thus ratio of MgSO4:H2O is (123.2-63.0)/120.4 : 63.0/18.0=0.5 : 3.5 =1 : 7
thus x=7