Answer: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of
b) moles of
According to stoichiometry :
1 mole of
require 1 mole of
Thus 0.0787 moles of
require=
of
Thus
is the limiting reagent as it limits the formation of product and
acts as the excess reagent. (10.0-0.0787)= 9.92 moles of
are left unreacted.
Mass of
Thus 19.9 g of
remains unreacted.
Answer:
this one is hard
Explanation:
but it's iron because the sodium so yea there u go.
Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT. Substituting a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
1. Convert 135g Fe2O3 to moles, divide by 2, and multiply the resulting number by the molar mass of Al