Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
Answer:
21.5 g.
Explanation:
Hello!
In this case, since the reaction between the given compounds is:

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

Now, the moles of Li3P consumed by 15 g of Al2O3:

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

Therefore, the total mass of products is:

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.
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