<h3> <u> Question 1</u></h3>
The number of grams of NO is 2.6 grams
<u><em>calculation</em></u>
4 NH₃ + 5O₂ → 4 NO +6H₂O
Step 1: find the moles of each reactant
moles =mass÷ molar mass
from periodic table the molar mass of NH₃ = 14 + (1 x3) = 17 g/mol
for O₂ = 16 x2 = 32 g/mol
moles for NH₃ =3.25 g÷ 17 g/mol =0.1912 moles
moles of O₂= 3.50 g÷ 32 g /mol=0.1094 moles
Step 2: Use the mole ratio to find the moles of NO
NH₃: NO is 4:4 = 1:1 therefore the moles of NO is also =0.1912 moles
O₂:NO is 5:4 therefore the moles of NO =0.1094 x 4/5 = 0.0875 moles
since O₂ yield less amount of NO, O₂ is the limiting reagent therefore the moles of No =0.0875 moles
Step 3: find the mass of No
mass = moles x molar mass
from periodic table the molar mass of No= 14 + 16 = 30 g/mol
mass =0.0875 moles x 30 g/mol = 2.6 grams
<h3> <u>question 2</u></h3>
The reactant which is limiting reagent is O₂
Explanation
4NH₃ + 5 O₂ →4 NO +6H₂O
Step 1: find the moles of each reactant
moles =mass÷ molar mass
from periodic table the molar mass of NH₃ = 17 g/mol, O₂ = 32 g/mol
moles of NH₃ = 3.25 g÷ 17 g/mol =0.1912 moles
moles of O₂ = 3.50 g÷ 32 g/mol=0.1094 moles
Step 2: use the mole ratio to determine the moles of product from each reactant.
from equation above NH₃:NO is 4:4= 1:1 therefore the moles of NO is also =0.1912 moles
O₂:NO is 5:4 therefore the moles of NO is =0.1094 moles x 4/5 =0.0875 moles
since O₂ yield less amount of NO , O₂ is the limiting reagent
<h3><u>
Question 3</u></h3>
The limiting reagent is CuCl₂
<u><em>calculation</em></u>
CuCl₂ + 2NaNO₃ → Cu(NO₃)₂ + 2 NaCl
Step 1: find the moles of each reactant
moles = mass÷ molar mass
from periodic table the molar mass of CuCl₂ = (63.5 + 35.5 x 2) =134.5 g/mol
for NaNO₃ = 23 + 14 +(16 x3) = 85 g/mol
moles for CuCl₂ = 15 g÷134.5 g/mol=0.1115 moles
moles of NaNo₃ = 20 g÷ 85 g/mol =0.2353 moles
Step 2: use the mole ratio to calculate the moles of the product from each reactant.
from equation above CUCl₂: NaCl is 1:2 therefore the moles of NaCl
=0.1115 moles x 2/1= 0.223 moles
NaNO₃: NaCl is 2:2 =1:1 therefore the moles of NaCl=0.2353 moles
since CUCl₂ yield less amount of NaCl, CuCl₂ is the limiting reagent
<h3><u>
Question 4</u></h3>
The grams of NaCl is 13.0 grams
<u><em>calculation</em></u>
CuCl₂ + 2NaNO₃ → Cu(NO₃)₂ + 2 NaCl
mass=moles x molar mass
from periodic table the molar mass of NaCl = 23 +35.5 = 58.5 g/mol
find the moles of NaCl as below
Step 1: find the moles of each reactant
moles = mass÷ molar mass
from periodic table the molar mass of CuCl₂ = (63.5 + 35.5 x 2) =134.5 g/mol
for NaNO₃ = 23 + 14 +(16 x3) = 85 g/mol
moles for CuCl₂ = 15 g÷134.5 g/mol=0.1115 moles
moles of NaNo₃ = 20 g÷ 85 g/mol =0.2353 moles
Step 2: use the mole ratio to calculate the moles of the product from each reactant.
from equation above CUCl₂: NaCl is 1:2 therefore the moles of NaCl
=0.1115 moles x 2/1= 0.223 moles
NaNO₃: NaCl is 2:2 =1:1 therefore the moles of NaCl=0.2353 moles
since CUCl₂ yield less amount of NaCl, CuCl₂ is the limiting reagent therefore the moles of NaCl =0.223 moles
Step 3: find moles of Nacl
= 0.223 moles x 58.5 g/mol =13.0 grams
