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.

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  • 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%