If 58.27 g ca react with 92.3 g hbr according to the reaction below, how many grams of hydrogen gas will be produced? unbalanced equation: ca + hbr → cabr2 + h2

 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

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.