a, mH=0,7g
mS=11,2g
mO=22,4g
b, mCa=5g
mC=1,5g
mƠ=6g
c, mBa=6,85g
mO=1,6g
mH=0,1g
d, mH=1g
mCl=35,5g
a) \(n_{H_2SO_4}=\dfrac{34,3}{98}=0,35\left(mol\right)\)
Ta có: \(n_H=2n_{H_2SO_4}=2\times0,35=0,7\left(mol\right)\)
\(\Rightarrow m_H=0,7\times1=0,7\left(g\right)\)
Ta có: \(n_S=n_{H_2SO_4}=0,35\left(mol\right)\)
\(\Rightarrow m_S=0,35\times32=11,2\left(g\right)\)
Ta có: \(n_O=4n_{H_2SO_4}=4\times0,35=1,4\left(mol\right)\)
\(\Rightarrow m_O=1,4\times16=22,4\left(g\right)\)
b)\(n_{CaCO_3}=\dfrac{12,5}{100}=0,125\left(mol\right)\)
Ta có: \(n_{Ca}=n_C=n_{CaCO_3}=0,125\left(mol\right)\)
\(\Rightarrow m_{Ca}=0,125\times40=5\left(g\right)\)
\(m_C=0,125\times12=1,5\left(g\right)\)
Ta có: \(n_O=3n_{CaCO_3}=3\times0,125=0,375\left(mol\right)\)
\(\Rightarrow m_O=0,375\times16=6\left(g\right)\)
c) \(n_{Ba\left(OH\right)_2}=\dfrac{8,55}{171}=0,05\left(mol\right)\)
Ta có: \(n_{Ba}=n_{Ba\left(OH\right)_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{Ba}=0,05\times137=6,85\left(g\right)\)
Ta có: \(n_O=n_H=2n_{Ba\left(OH\right)_2}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow m_O=0,1\times16=1,6\left(g\right)\)
\(m_H=0,1\times1=0,1\left(g\right)\)
d) \(n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\)
Ta có: \(n_H=n_{Cl}=n_{HCl}=1\left(mol\right)\)
\(\Rightarrow m_H=1\times1=1\left(g\right)\)
\(m_{Cl}=1\times35,5=35,5\left(g\right)\)
