Question

将氯化钠和碳酸钠的混合物33.8g放入125g稀盐酸中,恰好完全反应,得氯化钠溶液150g.计算

Answer 1

1.125g+33.8g-150g=8.8g(气体CO2的质量)

2HCl+Na2CO3=2NaCl+H2O+CO2

73 106 117 44

X Y 23.4g 8.8g

解得X=14.6g

Y=21.2g

2.14.6g/125g= 11.68%

150g-8.8g=141.2g

33.8g-21.2g=12.6g

12.6g+23.4g=36g

36g/8%=450g

450g-141.2g=308.8g