将表面氧化的镁带样品10.4g,浸入到10%的稀硫酸溶液中恰好完全溶解,并生成氢气0.2g.
1个回答

Mg + H2SO4 == MgSO4 + H2

24.98.120.2

m(Mg)..m1.m3.0.2g

所以24/m(Mg) = 98 / m1 = 120/m3 = 2/0.2

解得m(Mg) = 2.4g,m1 = 9.8g,m3 = 12g

m(MgO) = 10.4 - 2.4 = 8g

MgO + H2SO4 == MgSO4 + H2O

40.98.120

8g.m2.m4

所以40/8 = 98/m2 = 120/m4

解得m2 = 19.6g,m4 = 24g

所以稀硫酸的质量 = (m1 + m2)/10% = 294g

反应后溶液质量 = 10.4 + 294 - 0.2 = 304.2g

m(MgSO4) = m3 + m4 = 36g

所以反应后所得溶液的质量分数 = 36/304.2 * 100% = 11.8%

相关问题