Question

Bài1/ Tính khối lượng của:

a) 0,3mol kalisunfat

b) 0,4mol nhômclorua

c) 6,72 lít khí lưu huỳnhđioxit (ởđktc)

d) 11,2 lít khí oxi(ởđktc)

e) 12×1023 phân tử canxi nitat

f) 0,75 mol barihiđroxit

g) 10,5×1023 phân tử kẽmsunfat

h) 4,48 lít khí nitơ (ởđktc)

k) 0,8mol canxiclorua

l) 0,6 mol magiecacbonat

Answer 1

a) m\(_{K2SO4}=0,3.114=34,2\left(g\right)\)

b) m\(_{AlCl3}=0,4..133,4=53,4\left(g\right)\)

c)n\(_{SO2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

m\(_{SO2}=0,3.64=19,2\left(g\right)\)

d n\(_{O2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)

m\(_{O2}=0,5.32=16\left(g\right)\)

e) m\(_{Ba\left(OH\right)2}=0,75.171=128,25\left(g\right)\)

g) n\(_{ZnSO4}=\frac{10,5.10^{23}}{6.10^{23}0}=1,75\left(mol\right)\)

m\(_{ZnSO4}=1,75.161=281,75\left(g\right)\)

h)n\(_{N2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

m\(_{N2}=0,2.28=5,6\left(g\right)\)

k) m\(_{CaCl2}=0,8.111=88,8\left(g\right)\)

l) m\(_{MgCO3}=0,6.94=50,4\left(g\right)\)

Answer 2

https://i.imgur.com/IAxnnCC.jpg

Answer 3

https://i.imgur.com/pFMAG6s.jpg