Question

1. If 3.0 g of NH₃ reacts with 5.0 g of HCl, what is the limiting reactant?

NH₃(g) +  HCl(g) --> NH₄Cl(s)

2. How many grams of NH₄Cl would be produced in the previous problem?

0.137 g

7.32 g

24.6 g

53.46 g

Answer 1

O NH3(g) + HCl (g) → NHU CL IS) mass moles of 3 g 7=0.176 mol 1721 mol - molar mass- 1. of Hot Salmoj = 0.137 mol So, Hal is limiting reactant. moles of Whyce produced a 0.137 mol Now, massa molesy. molar mass of Nhy Cl = 0.137 mol X 53.5 g/mol - 17.329]