Asked by Pat
A 50g sample of ammonium dichromate is decomposed, what is the milliliter volume of nitrogen gas released at STP?
Answers
Answered by
bonjo
the balance equation;
(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g)
50g of ammonium dichromate is 0.198 mole (50/252.07). according to the equation, 1mole of ammonium dichromate produces 1 mole of nitrogen gas, so 0.198mole of dichromate produces 0.198mole of nitrogen gas.
mass of nitrogen gas = 0.198molex28g/mol = 5.54g
use the density of nitrogen gas to find the volume...
hope that helps..
(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g)
50g of ammonium dichromate is 0.198 mole (50/252.07). according to the equation, 1mole of ammonium dichromate produces 1 mole of nitrogen gas, so 0.198mole of dichromate produces 0.198mole of nitrogen gas.
mass of nitrogen gas = 0.198molex28g/mol = 5.54g
use the density of nitrogen gas to find the volume...
hope that helps..
Answered by
bonjo
if not, then recall that for any ideal gas at STP, 1mol=22.4L. and already we have the mole for nitrogen (0.198mole). so we can find the volume;
1mol=22.4L
0.198mol=x
x = 4.435L = 4435mL of Nitrogen gas.
1mol=22.4L
0.198mol=x
x = 4.435L = 4435mL of Nitrogen gas.
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