Question

In the unbalanced reaction, Sn + HNO3 ---> Sn(NO3)4+NO+H2O, 9.050 grams of Sn(NO3)4 was formed. How many NO molecules were formed and how many grams of HNO3 were reacted?

Answer 1

The balance recation is as follows :

3Sn + 16HNO₃ = 3 Sn(NO₃)₄ + 4NO + 8H₂O

Given that

Sn(NO₃)₄ = 9.050 g

Number of moles = amount in g/ molar mass

Moles of Sn(NO₃)₄ = 9.050 g/ 366.7296 g/ mole

= 0.025 moles Sn(NO₃)₄

stoichiometry of balance reaction, indicates the relationship between reactant and products of any chemical reaction.

Now calculate the moles of NO

0.025 moles Sn(NO₃)₄ *4 mole NO/3 moles Sn(NO₃)₄

0.033 Mole NO

1 Mole = 6.022*10^23 molecules

Therefore 0.033 Mole NO*6.022*10^23 molecules/1 Mole=0.2*10^23 molecules NO

Mole of HNO3

0.025 moles Sn(NO₃)₄ *16 mole HNO3/3 moles Sn(NO₃)₄

0.133 Mole HNO3

Amount in g= number of mole * molar mass

= 0.133 Mole HNO3 *63.01 g/mol

=8.40 g HNO3