Fc,0=65×0.34%=0.22(g) Ca0=65×0.52%=0.34(g) Mg0-65×0.47%=0.31(g) K20=65×0.53%=0.34(g) Na20=65x0.13%=10.085(g) 滑石中： si02-1×64.37%-0.64(g A10=1×1.27%:0.013(g) Fe0,-1x0.15%-0.0015(g) Ca0=13.3%=0.033(g) Mg0-1×30.91%-0.31(g) (3)将各原料中的同种氧化物加和，求出坯料中每种氧化物的总 质量： Si02=14.47+12.81+42.09+0.64-70.01(g) A1,0,-428+0.094+21.61+0.013-25.997(g) Fe0=0.16+0.035+0,22+0.0015-0,4165(g) Ca0-0.04+0.048+0.34-0.033=0.461(g) Mg0-0-0.033+0.31+0.31=0.653(g K,0=1.98+0+0.34+0=2.32(g) Na,0=1.07+0+0.085+0=1.155(g) (4)用每种氧化物的摩尔质量分别去除每种氧化物的质量，得到 每种氧化物的量n(mo1) Si02=70.01-60.1=1.1649(mo1) A1,0=25.997÷101.9=0.2551(m0o1) Fe0=0.4165-159.7-0.0026(mo1 Ca0=0.461+56.1=0.0082(mo1) Mg0-0.653+40.3=0.0162(mo1) K,0=2.32=94.2-0.0246(m01) Na0=l.155-62-0.0186(mol (5)计算出中性氧化物(RzO3)的总量： 0.2551+0.0026=0.2577(mo1) (6)以0.2577分别除每种氧化物的量