Question

Given the following reaction: 4 Zn + 10 HNO3 ? 4 Zn(NO3)2 + NH4NO3 + 3 H2O If we start with 3.32 g of Zn and excess of HNO3, how many g of each product will be formed?

Answer 1

*Determine number of moles of 1.29 g HNO3 using the molar mass of HNO3.

1.29 g HNO3 x (1 mole HNO3/63.0130 grams HNO3) = 0.0205 moles HNO3

*Determine the number of moles Zn that will react with 0.0205 moles HNO3.
Note: 10 moles of HNO3 reacts with 4 mole of Zn

0.0205 moles HNO3 x (4 moles Zn/10 moles HNO3) = 0.0082 moles Zn

*Convert 0.0082 moles Zn to atoms Zn using the Avogadro's number
Note: 1 mole Zn = 6.022 x 10^23 atoms Zn

0.0082 moles Zn x (6.022x 10^23 atoms Zn/1 mole Zn) = 4.94 x 10^21 atoms Zn