# Version A formular:

3Zn(OH)2 + 2Zn(CO)3(s)----------------------------->5ZnO(s)+2CO2(g) +3H2O(g)

105.4mL of Co2(g) was generated with a pressure of 765.5mm Hg for the gas. The temperature of the apparatus was 25.0C. How much zinc oxide would be left in tube if version A was the correct formula?

## Use PV = nRT and solve for n = mols CO2.

Then mols CO2 x (5 mols ZnO/2 mols CO2) = mols ZnO.

Then convert mols ZnO to g. g = mols x molar mass.

I would be somewhat concerned that CO2 is soluble in H2O, you haven't made a correction for the vapor pressure of H2O and you're classifying H2O as a gas at 25C.

## To determine the amount of zinc oxide left in the tube, we need to calculate the number of moles of CO2 generated and then use the stoichiometry of the reaction to find the number of moles of ZnO produced. Here's how you can do it step by step:

Step 1: Convert the pressure of CO2 gas from mmHg to atm

The given pressure is 765.5 mmHg. To convert it to atm, divide by 760 mmHg/atm:

765.5 mmHg ÷ 760 mmHg/atm = 1.0072 atm

Step 2: Convert the volume of CO2 gas to liters

The given volume is 105.4 mL. To convert it to liters, divide by 1000 mL/L:

105.4 mL ÷ 1000 mL/L = 0.1054 L

Step 3: Convert the temperature from Celsius to Kelvin

The given temperature is 25.0°C. To convert it to Kelvin, add 273.15:

25.0°C + 273.15 = 298.15 K

Step 4: Use the ideal gas law to calculate the number of moles of CO2

The ideal gas law equation is: PV = nRT

Where:

P = pressure in atm

V = volume in L

n = number of moles

R = ideal gas constant (0.0821 L·atm/(mol·K))

T = temperature in Kelvin

Plugging in the values:

(1.0072 atm)(0.1054 L) = n(0.0821 L·atm/(mol·K))(298.15 K)

Solving for n:

n = (1.0072 atm)(0.1054 L) / (0.0821 L·atm/(mol·K))(298.15 K)

n ≈ 0.0041 mol

Step 5: Use the stoichiometry of the reaction

From the balanced equation, we can see that the molar ratio of Zn(CO)3 to ZnO is 2:5.

Therefore, for every 2 moles of Zn(CO)3, we get 5 moles of ZnO.

Since the reaction produced 0.0041 moles of CO2, we can set up a ratio to find the moles of ZnO produced:

2 moles Zn(CO)3 : 5 moles ZnO = 0.0041 moles CO2 : x moles ZnO

Using this ratio:

x = (5 moles ZnO)(0.0041 moles CO2) / (2 moles Zn(CO)3)

x ≈ 0.0103 moles ZnO

Step 6: Convert moles of ZnO to grams

To convert moles of ZnO to grams, we need to know the molar mass of ZnO. The molar mass of ZnO is:

Zn: 65.38 g/mol

O: 16.00 g/mol

Molar mass of ZnO = (65.38 g/mol) + (16.00 g/mol) = 81.38 g/mol

Using the molar mass and the number of moles:

0.0103 moles ZnO × 81.38 g/mol = 0.838 g ZnO

Therefore, if version A is the correct formula, there would be approximately 0.838 grams of zinc oxide left in the tube.