Answer:
Grams of the reactant would be required is 8.26 g
Explanation:
Decomposition of NI3 is :
2 NI3 (s) --------------------> N2 (g)............... +..................... 3 I2 (g)
2 mol..................................1 mol.........................................3 mol
2 x 394.7 g/mol..............28 g/mol....................................3 x 253.8 g/mol
789.4 g...............................28 g...........................................761.4 g
From chemical equation it is cleared that
761.4 g of iodine is obtained from = 789.4 g of NI3
Therefore,
7.97 g of iodine is obtained from = 789.4 g x 7.97 g / 761.4 of NI3
= 6291.518 / 761.4 of NI3
= 8.26 g of NI3
Hence
Grams of the reactant would be required = 8.26 g
