<span>n BaSO4 = 11.65/233= 0.05</span> <span> n BaSo4 = n BaCO3 = 1\1</span> <span>m BaCO3 = 197 x 0.05 = 9.85г</span> <span> n BaCO3 = n CO2(во 2ом ур-ии) n CO2 = 0.05 => n CO2(общ) = 4.48/22.4 = 0.2 n CO2 в первом ур-ии = 0.2 - 0.05 = 0.15 n CO2(в 1ом ур-ии) = n Li2CO3 m Li2CO3 = 74 x 0.15 = 11.1 г n(CO2) = n H2SO4 в первом ур-ии = 1\1 m H2SO4 = 98 x 0.15 = 14.7 V CO2 = 0.15 x 22.4= 3.36 л w Li2CO3 = 11.1\ (11.1 + 14.7 - 3.36) x 100= 49.4\% => w BaCO3 = 100 - 49.4 = 50.5 \% </span>