The number of grams of hydrogen that will be produced is 1.142 grams
calculation
Step 1: write a balanced chemical equation
Ca + 2 HBr → CaBr₂ + H₂
Step 2: find the moles of each reactant
moles = mass /molar mass
The molar mass of Ca = 40 g/mol, for HBr = 1 + 79 .9 =80.9 g/mol
moles for Ca = 58.27 g /40 g/mol =1. 457 moles
moles for HBr = 92.3 g/ 80.9 g/mol =1.141 moles
Step 3: determine the limiting reactant
from equation above Ca :H2 is 1:1 therefore the moles of H₂ = 1.457 moles
HBr :H₂ is 2:1 therefore the moles of H₂ =1.141 moles x 1/2 =0.571 moles
since HBr produces least amount of H₂ it the limiting reactant and hence the moles of H₂ produced is 0.571 moles
Step 4: find the mass of H₂
mass = moles x molar mass
from periodic table the molar mass of H₂ = 2 g /mol
mass =0.571 x2 =1.142 grams
