Answer to Question #177060 in General Chemistry for devin

Question

1. Determine the mass of 3.011 x 1023 atoms of gold.
2. Calculate the number of formula units in 15.0 grams of CaSO4.
3. Calculate the number of moles in 168.0 g of HgS.
4. Which would have more molecules, 2.0 moles of hydrogen gas, H2, or 2.0 moles of methane gas, CH4? Explain your answer. Include any relevant calculations if needed to help support your answer.
5. Which would have the larger mass, 2.0 moles of hydrogen gas, H2, or 2.0 moles of methane gas, CH4? Explain your answer. Include any relevant calculations if needed to help support your answer.

Answer 1

Solution.

1.$N=3.011\sdot 10^{23};$

$M(Au)=197g/mol;$

$m_-?;$

$\nu=\dfrac{N}{N_A};$

$\nu=\dfrac{3.011\sdot10^{23}}{6.02\sdot10^{23}mol^{-1}}=0.5mol;$

$\nu=\dfrac{m}{M}\implies m=\nu M;$

$m=0.5mol\sdot197g/mol=98.5g.$

2.$m=15.0g;$

$M(CaSO_4)=136gmol^{-1};$

$N-?;$

$\nu=\dfrac{m}{M}=\dfrac{N}{N_A}\implies N=\dfrac{mN_A}{M};$

$N=\dfrac{15.0g\sdot 6.02\sdot 10^{23}mol^{-1}}{136gmol^{-1}}=6.64\sdot10^{22}.$

3.$m=168.0g;$

$M(HgS)=232.6gmol^{-1};$

$\nu-?;$

$\nu=\dfrac{m}{M};$

$\nu=\dfrac{168.0g}{232.6gmol^{-1}}=0.72mol;$

4.$\nu(H_2)=2mol;$

$\nu(CH_4)=2mol;$

$\nu=\dfrac{m}{M}=\dfrac{N}{N_A}\implies N=\dfrac{mN_A}{M}=\nu N_A;$

$N(H_2)=2mol\sdot6.02\sdot10^{23}mol^{-1}=12.04\sdot10^{23};$

$N(CH_4)=2mol\sdot6.02\sdot10^{23}mol^{-1}=12.04\sdot10^{23};$

5.$\nu(H_2)=2mol;$

$\nu(CH_4)=2mol;$

$M(H_2)=2gmol^{-1};$

$M(CH_4)=16gmol^{-1};$

$\nu=\dfrac{m}{M}\implies m=\nu M;$

$m(H_2)=2mol\sdot2gmol^{-1}=4g;$

$m(CH_4)=2mol\sdot16gmol^{-1}=32g;$

Answer: $1.$ $m=98.5g;$

$2.$ $N=6.64\sdot10^{22};$

$3.$ $\nu=0.72mol;$

$4.$ $N(H_2)=12.04\sdot10^{23}; N(CH_4)=12.04\sdot10^{23};$

$5.$ $m(H_2)=4g;m(CH_4)=32g.$