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?
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
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