The excess reactant when 4.35 g of hydrogen reacts with 30.75g of hydrogen iodide is hydrogen
calculation
write the equation for the reaction
H2(g) + I2 (g) → 2HI (g)
find the mole of each reactant
moles = mass/molar mass
moles of H2= 4.35 g/2 g/mol= 2.175 moles
moles o I2 = 30.75 g/ 254 g/mo=0.1211 moles
0.1211 moles of I2 reacted with 0.1211 moles of H2 but there are more moles of H2 there Hydrogen was in excess
