Answer:
Answer choice B
Explanation:
6Li + Ga₂S₃ => 3Li₂S + 2Ga
Given: 0.90 mole Ga Formed
Question: How many moles Li were used
Note => From the balanced equation note that 6 moles Li are required to produce 2 moles Ga. So, when mole ratios are applied, the value for moles of Lithium will be greater than the given value for the Ga yield. To take advantage of this idea stoichimetrically, use a ratio of the coefficients of Li and Ga as a multiple of the given to get a greater number of moles Li.
That is, use 6/2 x 0.90 mole => 2.7 moles Li used.
(Note: one would NOT want to use 2/6 x moles Ga b/c a number smaller than 0.9mole would be used.)
What I'm trying to share with you in this problem makes mole-mole problems very straight forward. When working chemical stoichiometry problems learn to use this rule => Convert all data to moles, apply mole ratio of given and unknown in the balanced equation to increase or decrease magnitude of the needed answer value.
Example
Given: N₂ + 3H₂ => 2NH₃
5moles ?moles H₂ are required
moles of H₂ required = 3/1(5) moles are required = 15moles H₂
Note, the ratio of coefficients chose to multiply times given moles is 3/1 because the value of H₂ must be larger than 5 moles b/c coefficient for H₂ is larger than coefficient for N₂. If 1/3 was chosen times the given 5 moles N₂ then the math gives 1/3(5) = 1.67 mole which is incorrect as it must be greater than 5 and the coefficients in the standard equation shows H₂ larger than N₂.
Using the 15 moles H₂ as the known reagent, how many moles NH₃ would be formed:
Note: In the balanced equation the coefficient for NH₃ is smaller than the coefficient for H₂. So, the answer must be smaller than 15; so use the mole coefficient ratio of 2/3 x known => a value smaller than the known and the correct answer. That is ...
moles NH₃ produced from 15 moles H₂ = 2/3(15) moles NH₃ formed = 10 moles NH₃ produced.
Note the moles NH₃ produced ( = 10 moles) is smaller than the given 15 moles of H₂ used.
