Question

mno2+4hcl=cl2+mncl2+2h2o if you were given 145 g of HCl how many grams of MnCl2 Could you Theoretically produce if you reacted with the excess MnO2

Answer 1

The first step is to use the molecular weight of HCl to convert 145g to moles of HCl (mw=36.458g/mol):

`145g\cdot(mol)/(36.458g)=3.97mol`

Now, use the ratio of the coefficients of MnCl2 to HCl, to find how many moles of MnCl2 are produced with this amount of HCl:

`3.97molHCl\cdot(1molMnCl_2)/(4molHCl)=0.9925molMnCl_2`

Use the molecular weight of MnCl2 to convert the amount of moles produced to grams:

`0.9925molMnCl_2\cdot(125.844g)/(molMnCl_2)=124.9g`

124.9g of MnCl2 are produced.