有5个网友回答
解：（1）反应过程中放出二氧化碳的质量为
84.8g-80.4g=4.4g
（2）设参加反应的碳酸钠为x。
Na2CO3+2HCl=2NaCl+H2O+CO2↑
106 44
x 4.4g
106∕x=44∕4.4g
x=10.6g
答：样品中碳酸钠的质量为10.6g。
（3）设参加反应的稀盐酸质量为y
Na2CO3+2HCl=2NaCl+H2O+CO2↑
73 44
y 4.4g
73∕y=44∕4.4g
y=7.3g
又稀盐酸溶液质量为73g
∴所用稀盐酸中溶质的质量分数为7.3g／73g×100％=10％
答：所用稀盐酸中溶质的质量分数为10％。
(1)84.8g-80.4g=4.4g
(2)解：设碳酸钠的质量为x,稀盐酸的质量为y.
Na2CO3+2HCl=NaCl+H2O+CO2↑
106 73 44
x y 4.4g
106 44
-------=-----
x 4.4g
x=10.6g
答：样品中碳酸钠的质量为10.6g
(3)73 44
----=----
y 4.4g
y=7.3g
7.3g
----*100%=10%
73g
答：稀盐酸溶质的质量分数为10%。
Na2CO3+2HCL==2NaCl+H2O+CO2(气体符号)
(1)4.4g(84.8g-80.4g)
(2)10.6g
(3)10%
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