\(n_{O_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(C+O_2\underrightarrow{t^O}CO_2\)
0,3-------------->(mol)
a) \(m_C=0,3.12=3,6\left(g\right)\)
\(m_{thanđá}=\frac{3,6}{95\%}=\frac{72}{19}\simeq3,789\left(g\right)\)
b) \(V_{CO_2}=0,3.22,4=6,72\left(l\right)\)
Chúc bạn học tốt
a) nO2 =\(\frac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH:
C+O2\(\underrightarrow{t^o}CO_2\)
0,3 ←0,3→0,3 (mol)
=> mC= 0,3.12=3,6(g)
=> mtđ= \(\frac{3,6}{95}.100=\frac{72}{19}\approx3,789\left(g\right)\)
b) VCO2= 0,3.22,4=6,72(l)
