Question

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Answer 1

Answer:

1) Mass = 18.864 g

2) V =  2.47 L

Explanation:

1) Given data:

Mass of water produced = ?

Mass of SnO₂ react = 79.0 g

Solution:

SnO₂ + 2H₂   →      Sn + 2H₂O

Number of moles of SnO₂:

Number of moles = mass/molar mass

Number of moles = 79.0 g/ 150.71 g/mol

Number of moles = 0.524mol

Now we will compare the moles of SnO₂ and H₂O

                 SnO₂         :        H₂O

                    1             :          2

                  0.524      :         2/1×0.524 =1.048 mol

Mass of water:

Mass = number of moles × molar mass

Mass = 1.048 mol × 18 g/mol

Mass = 18.864 g

2) Given data:

Volume of oxygen required = ?

Volume of C₃H₈ combust at STP = 0.499 L

Solution:

Chemical equation:

C₃H₈ + 5O₂     →      3CO₂ + 4H₂O

Number of moles of C₃H₈:

PV = nRT

R = general gas constant = 0.0821 atm.L /mol.K

1 atm × 0.499 L = n × 0.0821 atm.L /mol.K × 273.15 K

0.499 L atm.L = n × 22.43 atm.L /mol

n = 0.499 L atm.L / 22.43 atm.L /mol

n = 0.022 mol

Now we will compare the moles of propane and oxygen.

                  C₃H₈        :           O₂

                       1          :           5

                    0.022    :          5/1×0.022 = 0.11 mol

Volume of oxygen required:

PV = nRT

R = general gas constant = 0.0821 atm.L /mol.K

1 atm ×  V = 0.11 mol × 0.0821 atm.L /mol.K × 273.15 K

1 atm ×  V  = 2.47 atm.L

V = 2.47 atm.L  / 1 atm

V =  2.47 L