Answer:

Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32 60
CH₃OH + CO ⟶ CH₃COOH
m/g: 160
(a) Moles of CH₃OH

(b) Moles of CH₃COOH

(c) Mass of CH₃COOH

Answer:
See explanation
Explanation:
We have been told in the question that the equation of the reaction is; 1 slice of cheese + 2 slices of bread = 1 Grilled cheese sandwich ( mole ratio is, 1:2:1) .
Then the reagents are 10 slices of cheese 30 slices of bread. It then follows that 10 slices of cheese should be combined with 20 slices of bread according to the mole ratio.
However, we have 30 slices of bread and 10 slices of cheese so cheese is the limiting reactant while bread is the reactant in excess.
Yes, the number of glilled chese sandwishes he can make is decided by the limiting reactant because it gets used up most.
Object one is 5.2 g/cm3
object two is 3.46g/ml
There you go , This is my work i have did on that subject area