Question

已知5.6g的鐵和200g的稀硫酸恰好完全反應(yīng)．求：稀硫酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)．

Answer 1

考點(diǎn)：根據(jù)化學(xué)反應(yīng)方程式的計(jì)算,有關(guān)溶質(zhì)質(zhì)量分?jǐn)?shù)的簡(jiǎn)單計(jì)算

專題：溶質(zhì)質(zhì)量分?jǐn)?shù)與化學(xué)方程式相結(jié)合的計(jì)算

分析：欲求這種硫酸的溶質(zhì)質(zhì)量分?jǐn)?shù)，需先求出參加反應(yīng)的H₂SO₄的質(zhì)量，再根據(jù)溶質(zhì)質(zhì)量分?jǐn)?shù)=

溶質(zhì)質(zhì)量

溶液質(zhì)量

計(jì)算即可．

解答：解：設(shè)參加反應(yīng)的H₂SO₄的質(zhì)量為x
Fe+H₂SO₄=FeSO₄+H₂↑
56   98
5.6g x

56

5.6g

=

98

x

解得x=9.8g，
所以，該硫酸的質(zhì)量分?jǐn)?shù)為：

9.8g

200g

×100%=4.9%
答：這種硫酸的溶質(zhì)質(zhì)量分?jǐn)?shù)是4.9%．

點(diǎn)評(píng)：本題主要考查學(xué)生對(duì)物質(zhì)質(zhì)量分?jǐn)?shù)和溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算能力．學(xué)生需正確書寫出化學(xué)方程式，并能正確運(yùn)用溶質(zhì)質(zhì)量分?jǐn)?shù)公式進(jìn)行計(jì)算，才能正確答題．