# what are the answers to these stoichiometry equations? 1) How many atoms of chlorine are in 19.0 grams of MgCl2? 2) What is the volume of 2.60 moles of CO2(g) at STP? 3) What is the mass of 20.00...

what are the answers to these stoichiometry equations?

1) How many atoms of chlorine are in 19.0 grams of MgCl2?

2) What is the volume of 2.60 moles of CO2(g) at STP?

3) What is the mass of 20.00 liters of N2(g) if the temperature is 0 degrees Celsius and the pressure is 101.3 kPa?

4) How many grams of NaOH(s) would you need to prepare 500. mL of 0.160M NaOH(aq)

5) WHat volume of 12M HCl(aq) solution(concentrated hydrochloric acid) is needed to prepare 250. mL of 0.100M solution?

6) What is the resulting concentration of KCl when 100. mL of 0.250M KCl(aq) are added to 400. mL of water?

7) What volume of 0.360 M Mg(NO3)3(aq) contains 0.120 moles of NO3^-1 ions?

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**4) How many grams of NaOH(s) would you need to prepare 500. mL of 0.160M NaOH(aq)**

A) Volume = 500 ml = 0.5 lts

Molarity = 0.16 M

Molarity = moles /volume in lts

moles = molarity * volume = 0.16*0.5 = 0.08 moles.

Moles = mass/molar mass,

mass = moles * molar mass = `0.08*39.997` = 3.19976 g

3.19976 g of NaOH(s) would you need to prepare 500. mL of 0.160M NaOH(aq).

**5) WHat volume of 12M HCl(aq) solution(concentrated hydrochloric acid) is needed to prepare 250. mL of 0.100M solution?**

A) Volume of HCl V1 = ?

Molarity of HCl M1 = 12 M

Volume of solution V2 = 250 ml

Molarity of solution M2 = 0.100M

Formula `M_1*V_1 = M_2*V_2`

`V_1 = (M_2*V_2)/M_1`

`V_1 = (0.100*250)/12`

V1 = 2.083 ml

Therefore 2.083 ml of 12M HCl(aq) solution is needed to prepare 250. mL of 0.100M solution.

**6) What is the resulting concentration of KCl when 100. mL of 0.250M KCl(aq) are added to 400. mL of water?**

A) Volume of KCl V1 = 100 ml

Molarity of KCl M1 = 0.250 M

Volume of water V2 = 400 ml

Molarity of water M2 = ?

Formula `M_1*V_1 = M_2*V_2`

`M_2 = (M_1*V_1)/V_2`

`M_2 = (0.250*100)/400`

M2 = 0.0625 M of water.

0.0625 M is the resulting concentration of KCl when 100. mL of 0.250M KCl(aq) are added to 400. mL of water.

**7) What volume of 0.360 M Mg(NO3)3(aq) contains 0.120 moles of NO3^-1 ions?**

A) The actual formula is `Mg(NO_3)_2` .

1 mole of `Mg(NO_3)_2` will have 2 moles of `NO_3^-1` ions.

Therefore 0.120 moles of `NO_3^-1` will have `0.120/2` = 0.06 moles of `Mg(NO_3)_2`

Molarity = moles/Volume in litres

Volume in litres = Moles/molarity.

= `0.06/0.360`

= 0.1666 litres = 0.1666*1000 = 166.6 ml.

**1) How many atoms of chlorine are in 19.0 grams of MgCl2?**

A) Molar mass of `MgCl_2` = 95.2g/mol

95.2 g/mol contain 2 mole of Cl

Therefore 19 gm of `MgCl_2` will have (19*2)/95.2 = 0.399 mole.

1 Mole = `6.022 141 99 X 10^23` atoms

therefore 0.399 moles will have `(6.022 141 99 X 10^23)*3.99`

= `2.4028347*10^24` atoms.

Therefore 19.0 gm of MgCl2 have `2.4028347*10^24` atoms of Cl.

**2) What is the volume of 2.60 moles of CO2(g) at STP?**

A) At STP

Pressure (P) = 1 atm

Temperature (T) = 273 K

Gas constant R = 0.0820 L atm K−1 mol−1

Formula : PV=nRT

`V = (nRT)/P`

V = `(2.60*0.0820*273)/1`

V = volume = 58.20 litres

**3) What is the mass of 20.00 liters of N2(g) if the temperature is 0 degrees Celsius and the pressure is 101.3 kPa?**

A) Volume = 20 lts

Mass = ?

Temperature = 0+273 = 273 K

Pressure = 101.3 kpa

Gas constant R = 8.314 L kpa K−1 mol−1

PV = nRT

`n = (PV)/(RT)`

n = `(101.3*20)/(8.314*273)` = 0.8926 moles.

n = Moles = mass/molar mass

Mass = moles * molar mass = `0.8926*28.01` = 25.00 g

mass = 25 g