Question

From the balanced reaction below, when you have 3.33 moles of (NH4)2Cr2O7, how many grams of N2 will be produced

Answer 1

Answer:

$m_{N_2}=93.3gN_2$

Explanation:

Hello there!

In this case, for this stoichiometry-based problem, it is firstly necessary to realize that the decomposition of ammonium dichromate is given by:

$(NH_4)_2Cr_2O_7(s)\rightarrow N_2(g)+4H_2O(l)+Cr_2O_3(s)$

Thus, since the mole ratio between ammonium dichromate and the gaseous nitrogen (molar mass = 28.02 g/mol) is 1:1, we can compute the produced mass of the latter via stoichiometry as shown below:

$m_{N_2}=3.33mol(NH_4)_2Cr_2O_7*\frac{1molN_2}{1mol(NH_4)_2Cr_2O_7}*\frac{28.01gN_2}{1molN_2}\\\\ m_{N_2}=93.3gN_2$

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