1. Silicon dioxide (quartz) is usually unreactive but reacts with hydrogen fluoride according to the equation below. SiO2(s)+4HF(g) SiF4(g) + 2H2O(l)
Assume that 6.0 mol HF is added to 4.5 mol SiO2
a. What is the limiting reactant?
b. What is the excess reactant?
c. What is the maximum amount of SiFi4, in grams, that can be produced?
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Answer:
Limiting Reactant: 6.0 mol HF
Excess Reactant: 4.5 mol SiO2
Maximum amount of SiF4: 156 g
Explanation:
Ratio in the reaction is 1 mol SiO2 is to 4 mol HF. Since HF is greater in proportion, let's find how much of the SiO2 it needs.
Given 6 mol HF,
6 mol HF x (1 mol SiO2/4 mol HF) = 1.5 mol SiO2
We only need 1.5 mol SiO2, but we have 4.5 mol SiO2 in the given. Thus, SiO2 is the excess reactant.
To solve for the maximum amount of SiF4 in grams, we need to use the given limiting reactant and convert it to grams using its molar mass.
For molar mass, use periodic table:
Molar mass of SiF4 = 28 + (19)4 = 104 g/mol SiF4
Use to solve for the amount:
6 mol HF x (1 mol SiF4/4 molHF) x (104g SiF4/1 mol SiF4) = 156 g SiF4