Cho 5,4g Alo tác dụng hết 200ml H2SO4 a) Vh2=?(đkc) b) CMH2SO4 c) CMdd sau phản ứng 19/09/2021 Bởi Maya Cho 5,4g Alo tác dụng hết 200ml H2SO4 a) Vh2=?(đkc) b) CMH2SO4 c) CMdd sau phản ứng
$n_{Al}=\dfrac{5,4}{27}=0,2mol \\PTHH : \\2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2 \\a.Theo\ pt : \\n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3mol \\⇒V_{H_2}=0,3.22,4=6,72l \\b.Theo\ pt : \\n_{H_2SO_4}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3mol \\⇒C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5M \\c.Theo\ pt : \\n_{Al_2(SO_4)_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1mol \\⇒C_{M_{Al_2(SO_4)_3}}=\dfrac{0,1}{0,2}=0,5M$ Bình luận
Đáp án: \(a,\ V_{H_2}=6,72\ lít.\\ b,\ C_{M_{H_2SO_4}}=1,5\ M.\\ c,\ C_{M_{Al_2(SO_4)_3}}=0,5\ M.\) Giải thích các bước giải: \(a,\ PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑\\ n_{Al}=\dfrac{5,4}{27}=0,2\ mol.\\ Theo\ pt:\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\ mol.\\ ⇒V_{H_2}=0,3\times 22,4=6,72\ lít.\\ b,\ \text{Đổi 200 ml = 0,2 lít.}\\ Theo\ pt:\ n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,3\ mol.\\ ⇒C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5\ M.\\ c,\ Theo\ pt:\ n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,1\ mol.\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=0,2\ lít.\\ ⇒C_{M_{Al_2(SO_4)_3}}=\dfrac{0,1}{0,2}=0,5\ M.\) chúc bạn học tốt! Bình luận
$n_{Al}=\dfrac{5,4}{27}=0,2mol \\PTHH : \\2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2 \\a.Theo\ pt : \\n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3mol \\⇒V_{H_2}=0,3.22,4=6,72l \\b.Theo\ pt : \\n_{H_2SO_4}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3mol \\⇒C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5M \\c.Theo\ pt : \\n_{Al_2(SO_4)_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1mol \\⇒C_{M_{Al_2(SO_4)_3}}=\dfrac{0,1}{0,2}=0,5M$
Đáp án:
\(a,\ V_{H_2}=6,72\ lít.\\ b,\ C_{M_{H_2SO_4}}=1,5\ M.\\ c,\ C_{M_{Al_2(SO_4)_3}}=0,5\ M.\)
Giải thích các bước giải:
\(a,\ PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑\\ n_{Al}=\dfrac{5,4}{27}=0,2\ mol.\\ Theo\ pt:\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\ mol.\\ ⇒V_{H_2}=0,3\times 22,4=6,72\ lít.\\ b,\ \text{Đổi 200 ml = 0,2 lít.}\\ Theo\ pt:\ n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,3\ mol.\\ ⇒C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5\ M.\\ c,\ Theo\ pt:\ n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,1\ mol.\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=0,2\ lít.\\ ⇒C_{M_{Al_2(SO_4)_3}}=\dfrac{0,1}{0,2}=0,5\ M.\)
chúc bạn học tốt!