设碱粉样品中碳酸钠的质量为x,Na2CO3+CaCl2=CaCO3↓+2NaCl106 100x 5.0g 106 100 = x 5.0g ,解得x═5.3g所以样品中碳酸钠的质量分数为:5.3g/5.6g×100%═94.6%答:样品中碳酸钠的质量分数为94.6%.