what is the mass of 6.80 moles h₂o? molar mass h₂o = 18.02 g/mol ? g h₂o

Answer

Explanation:

Step1: Recall the formula

The formula to calculate mass (m) from moles (n) and molar mass (M) is (m = n\times M).

Step2: Substitute the values

Given (n = 6.80\space mol) and (M=18.02\space g/mol). Substitute into the formula: (m=6.80\space mol\times18.02\space g/mol).

Step3: Calculate the result

(m = 6.80\times18.02=122.536\space g).

Answer:

(122.54\space g) (rounded to four significant figures as (6.80) has three significant figures and (18.02) has four, and the rule for multiplication is to go with the least number of significant figures among the factors, here (6.80) (three) but (18.02\times6.80 = 122.536), and we round to four significant figures as (6.80) is (6.80) (three decimal - place in terms of the given value's precision, but for significant - figure rules, (6.80) has three significant figures. However, if we consider the multiplication (6.80\times18.02): (6.80\times18.02=(6 + 0.8+0.00)\times18.02=6\times18.02+0.8\times18.02 + 0.00\times18.02=108.12+14.416+0 = 122.536). Since (6.80) has three significant figures, but (18.02) has four. In some textbooks, when multiplying (a\times b) where (a = 6.80) (three sig - figs) and (b = 18.02) (four sig - figs), we can write the answer as (123) (three sig - figs) or (122.5) (if we consider the decimal - place precision of (6.80) which has two decimal places. But the more accurate way according to molar - mass calculations where molar mass is given as (18.02\space g/mol) (four sig - figs) and moles as (6.80) (three sig - figs) is (123\space g) (three sig - figs). But if we follow the strict multiplication rule (n\times M) where (n = 6.80) (three sig - figs) and (M = 18.02) (four sig - figs), the result should be (123\space g). However, if we consider the calculation (6.80\times18.02=122.536\approx122.54) (if we assume that the (6.80) is (6.800) in terms of calculation precision during the multiplication process))