a 0.96 millimole sample of Na2CO3 gave 0.9 millimoles of CO2 gas. If a 0.376 g of pure Na2CO3 was reacted with excess acid, what volume of gas will be measured on this apparatus

Na2CO3 + heat ==> Na2O + CO2

Therefore, 1 mol Na2CO3 = 1 mol CO2
Therefore, this apparatus has an efficiency of 0.9/0.96 = 0.9375.

The mols 0.376/molar mass = ?
mols CO2 = the same
Use PV = nRT and solve for volume, then v x 0.9375 since the process is not 100% efficient.

To find the volume of gas that will be measured on the apparatus, we need to use the given information about the moles of CO2 produced and the molar mass of Na2CO3 to calculate the number of moles of Na2CO3 reacted.

Step 1: Calculate the moles of CO2 produced
Given that 0.96 millimoles of Na2CO3 gave 0.9 millimoles of CO2 gas, we can deduce that the molar ratio between Na2CO3 and CO2 is 0.9/0.96.
Therefore, the moles of CO2 produced per mole of Na2CO3 is 0.9/0.96 = 0.9375.

Step 2: Calculate the moles of Na2CO3 reacted
Now, we can find the moles of Na2CO3 reacted using the moles of CO2 produced and the molar ratio.
Since the molar ratio between Na2CO3 and CO2 is 0.9:0.96, we can set up a proportion:
0.9375 mol CO2 / 0.96 mol Na2CO3 = x mol CO2 / 0.376 g Na2CO3

To solve for x, we rearrange the equation:
0.96 mol Na2CO3 * (0.9375 mol CO2 / 0.96 mol Na2CO3) = x mol CO2
x ≈ 0.9281 mol CO2

Step 3: Calculate the volume of gas
Finally, we can use the ideal gas law to calculate the volume of gas. The ideal gas law formula is:
PV = nRT

Where:
P = pressure (assumed to be constant)
V = volume of gas (what we need to find)
n = moles of gas (0.9281 mol CO2)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature

Since the pressure, temperature, and ideal gas constant are not given, we cannot calculate the exact volume. However, if you have the values for pressure and temperature, you can substitute them into the equation and solve for V.