Question

What is the maximum number of grams of SF4(g) that can be prepared from 6.00g of SCl2(g) and 3.50g of NaF(s)?
The equation for the reaction is:

3SCl2(g) + 4NaF(s) -> SF4(g) + S2Cl2 + 4NaCl(s)

Answer 1

the equation is already balanced (always check before doing any stoichiometry).

you have two reagents, the 6.00g SCl2 and the 3.50g NaF. we need to determine which one is the limiting reactant. so let's start by figuring out how many moles of SF4 can be made with 6.00g SCl2.

\[\frac{ 6.00g~SCl_{2} }{ }\frac{ 1~mol~SCl_{2} }{ 102.97g }\frac{ 1~mol~SF_{4} }{ 3~mol~SCl_{2} }= ? ~moles~SF_{4}\]

repeat this process and determine how many moles SF4 you can make with 3.5g NaF.

whichever one makes *less* SF4 is the limiting reactant. at this point you should know how many moles of SF4 you can make with the limiting reactant, so all that's left is to convert moles SF4 to grams SF4 by multiplying by the molar mass of SF4 (the molar mass of SF4 is 108.07 g/mol).

Answer 2

yes, that's correct. as some minor nitpicks, it would be best to avoid rounding until the very end if possible. as for sig figs, the final answer should only be rounded to 2 decimal places based on the quantities given in the problem.