Calculate AgCl, NaNO3 from 1.3g NaCl + 3.5g AgNO3

[Faiien]

Homework Statement

( 1.3 g NaCl )+( 3.5 g AgNO3) ---> (xg AgCl) + (yg NaNO3)

How many grams of each product result from the following reactions, and how many grams of which reactant is left over?

Enter your answers numerically separated by commas. Express your answer using two significant figures.
mNACl, mAgCl, mNaNO3 = g

Homework Equations

The Attempt at a Solution

1.3g NaCl*(1mole of NaCl/58.44g)=0.022 AgCl
3.5g AgNO3*(1mole of AgNO3/169.87g)=0.0206(limiting reagent)
0.0206 mols AgCl (143.32g/1 mol of AgCl)=3.0g
0.0206 mols NaNO3 (84.99/1 mol of NaNO3)=1.8g

I've figured out that AgNO3 is the limiting reactant which will yield 0.0206 moles of AgCl and NaNO3. I've calculated that 0.0206 moles of AgNO3 will yield 3.0g of AgCl and 1.8g of NaNO3.
This seems to be correct because 1.3g+3.5g = 3.0g+1.8g.
When I enter 1.3,3.0,1.8 into masteringchemistry, it tells me I'm incorrect and I don't understand why. I would really appreciate some help.


Doh! Solved it. Sorry everyone.

 

[Theodocious]

Dude I honestly made this account just to help you...
first off Mastering chem is a pain in the arse, further more, when you answered
the question you answered correctly, except for the Mass of NaCl
which is actually 9.2*10^-2 for some reason...

 

[Borek]

Dude I honestly made this account just to help you...

Year and half after the question was asked. Sigh.

Attached are results as calculated by EBAS stoichiometry calculator.