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Dmitry_Shevchenko [17]
3 years ago
5

Determine the mass of tin produced from 0.211 moles of hydrogen gas

Chemistry
1 answer:
max2010maxim [7]3 years ago
4 0
  The  mass  of   tin  produced  from  0.211  moles   of  hydrogen  is  calculated  as   follows

by writing  the  reaction  equation

that  is
  SnO2  +  2H2  ---->  Sn  +  2H2O

by  use  of  mole  ratio   the between  H2  to  Sn  which  is   2:1

the  moles  of  Sn   is  therefore = 0.211/2=0.1055  moles

mass  =  moles  x  molar  mass  of  Sn

that  is  0.1050   x   118.71 grams



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Answer:

3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-

Explanation:

Hello there!

In this case, according to the required half-reaction, we start by setting it up from bismuth (V) oxide ion to bismuth (III) ion:

BiO_3^-\rightarrow Bi^{3+}

Thus, next realize that the oxidation state of Bi in BiO3^- is 5+ because oxygen is 2- (-2*3+x=-1;x=-1+6;x=+5), so we obtain:

(Bi^{5+}O_3)^-\rightarrow Bi^{3+}

Thereafter, we realize three water molecules are needed on the right in order to balance the oxygens and consequently 6 hydrogen atoms on the left to balance hydrogen:

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Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:

6OH^-+6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O-3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-

Then, we accommodate the waters to obtain:

3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-

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