can somebody help me with my chem homework please?

Question

Nitroglycerin, an explosive compound, decomposes according to the equation below.
4 C3H5(NO3)3(s) 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)

Calculate the total volume of gases when collected at 1.2 atm and 25°C from 1.9 102 g of nitroglycerin.

Molar mass of nitroglycerin is 227.0872

Answer 1

We need the complete equation.

Answer 2

4 C3H5(NO3)3(s) ----------> 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)

Moles of nitroglycerin = 1.9102 g / 227.0872 g/mole = .0084117 moles

Moles of Gases produced from 4 moles of nitroglycerin:

CO2 =12 moles
H2O = 10 moles
N2 = 6 moles
O2 = 1mole

Total = 29 moles of gases

Moles of gases produced from one mole nitroglycerin = 29 / 4 = 7.25 mole

Moles of gases produced from 1.9102 grams:

7.25 X .0084117 moles = 0.060985 moles

V = nRT/ P = (0.060985 moles)(.0.0821 l atm/mole K)(298 K) / 1.2 atm

V = 1.24 liters

Answer: 1.2 liters (limited to 2 significant figures because of "1.2 atm" and "25 C."

Answer 3

The number of moles of nitroglycerin which decomposes is found by dividing the mass of nitroglycerin by the molar mass.
= 1.9102/227.0872
=0.0084 moles
4 moles of nitroglycerin decomposes into 29 moles of gas, so 0.0084 moles will decompose into 0.0609 moles of gas.

To calculate the volume this gas will take up in the given conditions, we will use the ideal gas law, PV = nRT.
(pressure x volume = moles of gas x gas constant x temperature (K))
Rearranged, this gives V = nRT/P.
V = 0.0609 x 0.0821 x 298/1.2
V = 1.4899/1.2
V = 1.2416 L
1.9102 grams of nitroglycerin will decompose to produce 1.2416 litres of gas at 1.2 atm pressure and 25°C.

Answer 4

1.9 102 g, I'm assuming to be 1.9 x 10² = 190g ??

Moles of Nitro decomposed…
= 190g/227.0872g/mole
= 0.8367 moles
4 moles of Nitro produces 29 moles of gas, therefore, 0.8367 moles will decompose into …
29/4 x 0.8367 = 6.07 moles of gas.

Using the Ideal Gas Law…
PV = nRT
V = nRT/P
V = (6.07 x 0.0821 x 298) ÷ 1.2
V = 123.76L

Using the Combined Gas Law…
1 mole = 22.4L at STP
6.07moles = 22.4 x 6.07 = 136L
1.2atm x V1 x 273K = 1atm x 136L x 298K
V1 = 136 x 298 ÷ 1.2 x 273
V1 = 40,528 ÷ 327.6 = 123.71 L.