Câu hỏi của Hải Yến

Question

1.Cân bằng phương trình

Na+O2-----Na2O

Na2PO4+BaCL2------NaCl+Ba2(PO4)2

Al2O3+H2SO4-----AL2(SO4)3+H2O

2.Cho a gam nhôm(AL)tác dụng vừa đủ với dung dịch 29,4 gam axit sunfiric (H2SO4).Sau phản ứng...

Nguyễn Việt Hoàng · Accepted Answer

1/ $4Na+O_2\underrightarrow{t^o}2Na_2O$

$2Na_2PO_4+2BaCl_2\rightarrow4NaCl+Ba_2\left(PO_4\right)_2$

$Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O$

2/ $2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2$

$n_{H_2SO_4}=\frac{m}{M}=\frac{29,4}{98}=0,3mol$

$0,2-0,3-0,1-0,3mol$

$m_{Al}=n.M=0,2.27=5,4g$

$V_{H_2}=n.22,4=0,3.22,4=6,72l$Câu trả lời: 1/ $4Na+O_2\underrightarrow{t^o}2Na_2O$

$2Na_2PO_4+2BaCl_2\rightarrow4NaCl+Ba_2\left(PO_4\right)_2$

$Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O$

2/ $2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2$

$n_{H_2SO_4}=\frac{m}{M}=\frac{29,4}{98}=0,3mol$

$0,2-0,3-0,1-0,3mol$

$m_{Al}=n.M=0,2.27=5,4g$

$V_{H_2}=n.22,4=0,3.22,4=6,72l$

Nguyễn Duy Khang · Answer

1.

4Na+O2----->2Na2O

2Na2PO4+2BaCL2------>4NaCl+Ba2(PO4)2

Al2O3+3H2SO4-----AL2(SO4)3+3H2O

2.

a)nH2SO4=29,4/98=0,3mol

$2Al+3H2SO4\rightarrow Al2\left(SO4\right)3+3H2$

0,2_____0,3________0,1_________0,3(mol)

b) $a=0,2.27=5,4\left(g\right)$

c) $V_{H2}=0,3.22,4=6,72\left(l\right)$Câu trả lời: 1.