How many moles of lithium nitrate are theoretically produced if we start with 3.0 moles of Ca(NO3)2 and 3.0 moles of Li3PO4? Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2

6.0 mol LiNO₃

Explanation:

First we need to determine which reactant is the limiting reactant:

We calculate with how many Ca(NO₃)₂ moles would 3.0 moles of Li₃PO₄ react, using the stoichiometric coefficients of the reaction:

There are not as many Ca(NO₃)₂ moles, so Ca(NO₃)₂ is the limiting reactant.

Using the number of moles of the limiting reactant, we calculate how many moles of LiNO₃ would be formed:

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

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