Question

as vrea si eu va rog 5 probeme la chimie pt clasa a 8

Answer 1

1) 200g sol. hidroxid de sodiu,de concetratie 20% se neutralizeaza cu 400g sol. acid clorhidric,de concetratie 18,25%.Identificati  substanta  care este in exces si aflati masa acesteia.Calculati masa de sare care se obtine
R: ms1:200g NaOH
C1:20%
ms2:400g HCl
C2:18,25%

ms=?g exces

Na OH+H Cl=NaCl+H2O
C1=md1/ms1 ×100
20=md/200g ×100
md=40gNaOH
C2=md2/ms2 ×100
md=18,25×4=73g HCl
MNaOH=23+16+1=40g
MHCl=1+35,5=36,5 g
MNa Cl=23+35,5=58,5 g
40/40=58,5/x⇒x=40×58,5/40=58,5gNa Cl
Mexces=1×36,5=36,5g HCl

2)m=36 kg carbune
C=80%
?moli CO2

R:  C+O2=Co2
100kg carbune........80kg C
36kgcarbune......x kg C
x=36×80/100=144/5=28,8Kg C
12kg/28,8kg=1 mol/x moli⇒x=28,8×1/12=2,4moli CO2

3)m=100g ciocolata
m1=3,6g MgCu
m2=1,8MgCu/zi
R:100g cioco.......2,6 MgCu
xg cioco.........1,8 MgCu/zi
x=100×1,8/3,6=50g cioco

4)xgHCl.....x=10%......0,65g Zn
2 HCl+Zn=ZnCl2+H2↑
x=2×36,5×0,65/6,5=7,3g HCl pur
10/100=7,3/m1⇒m1=7,3×100/10=73g HCl

5)ms=8,9g H2SO4
C=10%
?g Mg
R:Mg+H2SO4=MgSO4+H2↑
C=md/ms×100
md=8,9×100/10
10/100=md/8,9
md=8,9×10/100=8,9/10=0,89

Sper ca sunt de folos,si ca nu le-ati facut