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2HNO₃+ Ba(OH)₂---> Ba(NO₃)₂ + 2H₂O
20g xg
nasytepnie liczysz M
HNO₃=1+3*16+14=63g/mol
Ba(NO₃)₂= 137+2*14+6*16=261g/mol
nastepnie ukladasz proporcje
20gHNO₃-----x g soli
2*63g(126)---261 g soli
x=41,43 g soli