Need on some homework questions i’m stuck on about molarity and dilutions:

1. how many moles of c12h22o11 (sucrose) are in 7.5 l of a 5.8 m c12h22o11 solution?

2. how many moles of solute licl are in 375 grams of an 85.0 percent by mass solution?

3. how many grams of copper metal will react completely with 4.50 l of 6.00 m hcl?
cu (s) + 2hcl (i) -> cucl2 +h2 (g)

4. completely describe how to make 3.5 l of a 2.0 m nacl solution from a stock solution of 6.0 m nacl.

5. how many grams of a stock solution that is 90.0% hcl by mass would be needed to make 175 grams of a 40.0% hcl by mass solution?

6. how many grams of silver metal would you expect to be produced when 250. grams of an 80.0% by mass silver nitrate solution reacts with excess copper metal?
cu (s) 2agno3 (aq) -> 2ag (s) + cu(no3)2 (aq)

you so much and try to answer all of the questions.

  question 1

The  moles of C12H22O11  that  are in  7.5 L  of a  5.8 M C12H22O11 solution  is    43.5  moles

 calculation

moles = molarity  x volume

volume = 7.5 L

molarity = 5.8 M or  5.8 mol/l

moles is therefore = 7.5 l x 5.8 mol/l = 43.5 moles

question 2

% percent  =( mas of the solute/ mass of the solution)  x 100

mass of solute= ?

mass of the solution   = 375 g

% mass =  85 %  = 85/100

let the mass of solute be represented by Y

therefore  y/375  = 85/100

y=(375 x 85)/100 = 318.75 g

moles =  mass/molar mass

= 318.75 g/ 42.4= 7.52  moles

Question  3

 The grams of copper  metal is  857.25 g

 calculate the  moles  HCl

=  molarity x volume  = 4.50 L x 6 .00 mol/l= 27 moles

  By use of mole ratio of Cu: HCl which is  1:2 the moles of Cu = 27/2 =13.5 moles

mass  = moles  x molar mass

= 13.5  moles x 63.5  g/mol = 857.25 g

question 4

Complete description of how to make  3.5 L of 2.0 M NaCl solution  from a stock of 6.0 M NaCl  is   that measure  1.17 L  of 6.0 M NaCl, Dissolve it  to 3.5 L to make 2.0 M NaCl

calculation

M1V1 = M2V2

m1=  2.0 M

V1=3.5 L

M2= 6.0 M

V2=?

v2  = M1V1/M2

= (3.5 L x  2.0 M)/ 6.0 M= 1.17 L

explanation : measure  1.17  L of 6.0M NaCl  and then dissolve  it to 3.5 L  to make 2.0 M NaCl

question 5

The number of grams  of stock  solution that is in 90.0 %  is HCl by mass  that would be  needed  to  make 175 g is 77.78 grams

calculation

This  is calculated using M1C1= M2C2 formula

M1  = ?g

C1  =90.0%

M1= 175 g

C1= 40.0 %

make M1 the subject of the  formula

M1= M2C2/C1

M1 is  therefore =( 175 g x 40) / 90 = 77.78 g

Question  6

The grams  of silver metal is 127 g

% mass =  mass of solute/ mass of solution

% mass  = 80% = 80/100

mass of solute (AgNO3)= ?

mass  of solution = 250 g

let the mass of solute be  represented by Y

therefore  Y/ 250 = 80/100

Y= (250 x 80) /100 =200 g  of  AgNO3

moles = mass/molar mass

moles of AgNO3 =  200 g/ 169.87 g/mol =1.178  moles

The mole ratio of AgNO3: Ag is 2:2=1:1  therefore the moles of Ag= 1.178 moles

mass=  moles  x molar mass

= 1.178  moles x  107.87 g/mol =127.07 g

the millers tale is the answer.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3