Đáp án:
Giải thích các bước giải:
a)Gọi $n_{MO}=a(mol)$
$MO+H_2SO_4→MSO_4+H_2↑$
$a:a:a:a$
$m_{H_2SO_4}=98a(g)$
$m_{ddH_2SO_4}=98a:4,9$%$=2000a(g)$
$m_{ddMSO_4}=aM+2000a-2a=aM+1998a(g)$
$C_{MSO_4}=\frac{a(M+96)}{a(M+16)+1998a}.100=7,69$
$⇔100aM+9600a=7,69aM+123,04a+15364,62a$
$⇔92,31aM=5887,66a$
$⇔\frac{aM}a=\frac{5887,66}{92,31}≈64$
$⇔M=64(Cu)$
b)$V=\frac{2000a}{1,86}≈1075,269aml≈1,075a(l)$
Xin hay nhất!!!