3NO2(g)+H2O(l)2HNO3(l)+NO(g) A) Suppose that 4.4 mol NO2 and 0.60 mol H2O combin

Question

3NO2(g)+H2O(l)2HNO3(l)+NO(g)

A) Suppose that 4.4 mol NO2 and 0.60 mol H2O combine and react completely. Which reactant is in excess?
B) How many moles of the reactant in excess are present after the reaction has completed?

2Na3PO4(aq)+3CuCl2(aq) Cu3(PO4)2(s)+6NaCl(aq)

C) What volume of 0.194 M Na3PO4 solution is necessary to completely react with 90.0 mL of 0.116 M CuCl2?

Calculate the molarity of each of the following solutions

D) 5.15 mol of LiCl in 2.90 L solution ______ M
E) 27.43 g C6H12O6 in 1.16 L of solution _____ M
F) 36.4 mg NaCl in 127.1 mL of solution ____M

I'm having a lot of trouble with these questions!!! :(

Explanation / Answer

3NO2(g)+H2O(l)2HNO3(l)+NO(g)

A) Suppose that 4.4 mol NO2 and 0.60 mol H2O combine and react completely. Which reactant is in excess?

Here NO2 reactant is in excess; H2O reacted completely .

0.60 mol H2O * 3 mole NO2/ 1 Mole H2O

= 1.8 mole NO2

B) How many moles of the reactant in excess are present after the reaction has completed?

Excess mole of NO2 = 4.4 Mole – 1.8 mole

= 2.6 mole NO2

2Na3PO4(aq)+3CuCl2(aq) Cu3(PO4)2(s)+6NaCl(aq)

C) What volume of 0.194 M Na3PO4 solution is necessary to completely react with 90.0 mL of 0.116 M CuCl2?

First calculate the moles of CuCl2:

Number of moles = molarity * volume in L

= 0.116 M CuCl2 * 90/1000

= 0.01044 Moles CuCl2

Now calculate the number of moles of Na3PO4 as follows:

0.01044 Moles CuCl2* 2 mole Na3PO4 /3.0 Moles CuCl2 = 0.00696 moles of Na3PO4

Molarity = number of moles / volume in L

SO;

Volume in L = number of moles / molarity

= 0.00696 moles of Na3PO4 /0.194 M

= 0.0359 L

= 35.9 ml

Calculate the molarity of each of the following solutions

D) 5.15 mol of LiCl in 2.90 L solution ______ M

Molarity = number of moles / volume in L

= 5.15 mol of LiCl / 2.90 L

= 1.78 M

E) 27.43 g C6H12O6 in 1.16 L of solution _____ M

27.43 g C6H12O6 in 1.16 L of solution = ( 27.43 g / 180.16 g·mol1 ) / 116 L = 0.131 M

F) 36.4 mg NaCl in 127.1 mL of solution ____M

36.4 mg NaCl in 127.1 mL of solution = ( 36.4 x 10^-3 g / 58.44 g/mol ) / 127.1 x 10^-3 L = 0.0049 M

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