(2008?南岗区二模)实验室有一瓶浓硫酸.瓶上标签内容如右图所示,请根据标签上提供的数据解答下列问题 (2008?南岗区一模)如右图所示,是物质发生化学变化的微观...
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(1)设所需浓硫酸的质量为y,2NaOH+H2SO4=Na2SO4+2H2O
80 98
80g 98%x
∴
80 |
98 |
80g |
98%y |
解之得:y=100g.
故答案为:100g.
(2)①M和硫酸反应的方程式为:M+H2SO4=MSO4+H2↑.
②设反应生成氢气质量为x,M的相对原子质量为z,
M+H2SO4=MSO4+H2↑
z 98 2
10g-7.6g 98g×10% x
∴98:2=(98g×10% ):x,z:98=(10g-7.6g ):(98g×10%)
解之得:x=0.2g,z=24.
③由②中计算出的M的相对原子质量为24可知,此金属是镁,盐酸符号为Mg.
④合金中铜元素的质量分数为:
7.6g |
10g |
⑤根据质量守恒定律,反应生成的MgSO4的质量为:(10g-7.6g)+(98g×10%)-0.2g=12g,
所得溶液的质量为:98g+10g-7.6g-0.2g=100.2g
所得溶液溶质的质量分数为:
12g |
100.2g |
故答案为:①M+H2SO4=MSO4+H2↑;②98:2=(98g×10% ):x;③Mg;④76%;⑤12.0%.
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