can't really figure /sort this one out

so the given balenched equation

Na2SiO3 + 8 HF  ----> H2SiF6 + 2 NaF + 3H2O

how many grams of NaF form when 0.500 mol of HF react with excess Na2Sio3 ?

I think it's a limiting reagent problem someplace.. but not sure plz help asap

From equation
8moles of HF give 2 moles of NaF
Now number of moles of HF= 0.500
No of moles of NaF formed = 0.500  X 2/ 8 = 0.125
I have used the factor 2/8 to convert it to moles of NaF
Now you just need to multiply moles of NaF with its molarmass to get the answer.