①在标准状况下,112mL某气体的质量是0.14g,则其相对分子质量为______.②含有相同原子个数的SO2和SO3,其质量比为______,摩尔质量比为______,③200mL2.00mol/L

题目简介

①在标准状况下,112mL某气体的质量是0.14g,则其相对分子质量为______.②含有相同原子个数的SO2和SO3,其质量比为______,摩尔质量比为______,③200mL2.00mol/L

题目详情

①在标准状况下,112mL某气体的质量是0.14g,则其相对分子质量为______.
②含有相同原子个数的SO2和SO3,其质量比为______,摩尔质量比为______,
③200mL2.00mol/L的Al2(SO43溶液中含有______mol Al2(SO43,Al3+的物质的量浓度为______mol•L-1,含Al3+的质量为______g.
题型:问答题难度:中档来源:不详

答案

①n=class="stub"0.112L
22.4L/mol
=0.005mol,M=class="stub"m
n
=class="stub"0.14g
0.005mol
=28g/mol,则相对分子质量为28,故答案为:28;
②假设原子个数为N,则n(SO2)=class="stub"1
3
×class="stub"N
NA
mol,n(SO3)=class="stub"1
4
×class="stub"N
NA
mol,
m(SO2):m(SO3)=class="stub"1
3
×class="stub"N
NA
mol×64g/mol:class="stub"1
4
×class="stub"N
NA
mol×80g/mol=16:15;
M(SO2):M(SO3)=64g/mol:80g/mol=4:5,
故答案为:16:15;4:5;
③n(Al2(SO4)3)=0.2L×2.00mol/L=0.40mol;
n(Al3+)=2n(Al2(SO4)3)=0.80mol,
c(Al3+)=2c(Al2(SO4)3)=4.00mol/L;
m(Al3+)=0.80mol×27g/mol=21.6g,
故答案为:0.40;4.00;21.6.

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