4questions!
4nh3 + 5o2 4no +6h2o

in an experiment, 3.25 g of nh3 are allowed to react with 3.50 g of o2. how many grams of no are formed?

round final answers to the tenth position, one place after the decimal. only report the numerical answer.

question 2 (25 points)
question 2 options:
take the reaction: nh3 + o2 no + h2o. in an experiment, 3.25 g of nh3 are allowed to react with 3.50 g of o2. which reactant is the limiting reagent?
you may either write the formula or the name of the limiting reactant.
question 3 (25 points)
question 3 options:
cucl2 + 2 nano3 > cu(no3)2 + 2 nacl

if 15 grams of copper (ii) chloride react with 20 grams of sodium nitrate, which is the limiting reagent?

you may either write the formula or the name of the limiting reactant.
question 4 (25 points)
question 4 options:
cucl2 + 2 nano3 > cu(no3)2 + 2 nacl

if 15 grams of copper (ii) chloride react with 20 grams of sodium nitrate, how many grams of nacl is formed?

round final answers to the tenth position, one place after the decimal. only report the numerical answer.

The number of  grams of NO  is   2.6 grams

calculation

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

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

The limiting reagent  is  CuCl₂

 calculation

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

The  grams of NaCl  is   13.0 grams

 calculation

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

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.