1)
BaCO3 = BaO + CO2
n(BaCO3) = m/Mr = 19,7/197 = 0,1 моль
n(BaO) =n(BaCO3) = 0,1 моль
m(BaO) = n*Mr = 0,1* 153 = 15,3 г
V(CO2) = Vm* n = 22,4*0,1 = 2,24л
Ответ 15,3 г и 2,24 л
2)
2N2 + 3H2 =2 NH3
<span>28кг<span> </span>х кг теоритически</span>
<span><span><span> </span>N2</span></span><span> <span> </span>+ <span> </span>3 Н2 <span> </span>= <span> </span><span>2 NН3</span></span>
<span>28 кг<span> </span>2·17кг</span>
<span> </span>
<span>m (NН3)теоретич =(28·2·17):28=34кг</span>
<span>выход продукта= mпр : mтеор · 100% = 25,5 : 34 ·100 =75%</span>
<span>3)</span>
<span>CO2 + Ca(OH)2 = CaCO3 + H2O</span>
<span>n(CO2) = V/Vm = 6,72/22,4 = 0,3 моль</span>
<span>n(CaCO3) = 0,3 моль</span>
<span>m(CaCO3) = n*Mr = 0,3* 100 = 30 г</span>
<span>Ответ 30 г. </span>
<span>4)Fe2O3 + H2 = Fe + H2O.</span>
<span> 1)m(Fe2O3) = 16*0,8 = 12,8 г</span>
<span>2) n(Fe2O3) = m/Mr = 12,8/160 = 0,08 моль</span>
<span>3) n(Fe) = 0,08 моль</span>
<span>4)m(Fe) = 0,08*56 = 4,48г</span>
<span>Ответ: 4,48 г</span>
<span>5)</span>
<span>MgO+ H2SO4 = MgSO4 + H2O</span>
<span>m(H2SO4) = 49*0,1 = 4,9 Г</span>
<span>n(H2SO4) = m/Mr = 4,9/98 = 0,05 моль</span>
<span>n(MgSO4) = 0,05 моль</span>
<span>m(MgSO4) = n*Mr = 0,05 * 120 = 6 </span>
<span>Ответ: 6 г.
</span>