Đáp án:
32,98%
Giải thích các bước giải:
\(SO_3+H_2O\to H_2SO_4\)
\(n_{SO_3}=\dfrac{200}{80}=2,5\ \text{mol}; m_{dd\ H_2SO_4}=1000\cdot 1,12=1120\ \text{gam}\to m_{H_2SO_4}=1120\cdot 17\%=190,4\ \text{gam}; m_{H_2O}=1120-190,4=929,6\ \text{gam}\to n_{H_2O}=\dfrac{929,6}{18}=\dfrac{2324}{45}\ \text{mol}\)
Ta có: \(2,5<\dfrac{2324}{45}\to\)Sau phản ứng \(SO_3\) hết
\(\to n_{H_2SO_4}=n_{SO_3}=2,5\ \text{mol}\to m_{H_2SO_4}=98\cdot 2,5=245\ \text{gam}\)
BTKL: \(m_{dd\ A}=200+1120=1320\ \text{gam}\)
\(\to C\%_{dd\ A}=\dfrac{245+190,4}{1320}\cdot 100\%=32,98%\%\)
