дано
m(ppa Ba(NO3)2) - 2.61 g
W(Ba(NO3)2) = 10%
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m(BaSO4)-?
m(Ba(NO3)2) = 2.61 * 10% / 100% = 0.261 g
K2SO4+Ba(NO3)2-->2KNO3+BaSO4
M(Ba(NO3)2) = 261 g/mol
n(Ba(NO3)2) = m/M =0.261 / 261 = 0.001 mol
n(Ba(NO3)2) = n(BaSO4) = 0.001 mol
M(BaSO4) = 233 g/mol
m(BaSO4) = n*M = 0.001 * 233 = 0.233 g
ответ 0.233 г