Part 1:

Yes, every letter only appears once in each column and row hence it is.

Part 2:

The linear equation would be:

\[y_{ijk} = \mu + \tau_i +\beta_j + \gamma_k + \epsilon_{ijk} \]

where \(\mu\) is the overall mean, \(\alpha_i\) is the factor of treatments, Both \(\beta_j\) and \(\gamma_k\) are blocks and \(\epsilon_{ijk}\) is the random error.

observation <- c(8,7,1,7,3,11,2,7, 3, 8, 4, 9, 10, 1, 5, 6, 8, 6, 6, 10, 4, 2, 3, 8, 8)
batch <- factor(rep(1:5, each = 5))
day <- c(rep(seq(1:5),5))
ingredient <- c('A', 'B', 'D', 'C', 'E', 'C', 'E', 'A', 'D', 'B', 'B', 'A', 'C', 'E', 'D', 'D', 'C', 'E', 'B', 'A', 'E', 'D', 'B', 'A', 'C') 
day <- as.factor(day)
ingredient <- as.factor(ingredient)
data <- data.frame(observation, batch, day, ingredient)
aov.model <- aov(observation~batch+day+ingredient, data=data)
summary(aov.model)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## batch        4  15.44    3.86   1.235 0.347618    
## day          4  12.24    3.06   0.979 0.455014    
## ingredient   4 141.44   35.36  11.309 0.000488 ***
## Residuals   12  37.52    3.13                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The p-value of ingredient is 0.000488 is less than 0.05, so we reject the NULL hypothesis and accept that there is a significant effect of the ingredients in the reaction time.