Question

How many grams are present in 3.4
moles of CH2COOH?
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Answer 1

m = 200.6 gr

Step-by-step explanation:

CH2COOH : n = 3.4 mol

Mr = 12 + 2 + 12 + 2×16 + 1 = 59

n= m / Mr => m = n× Mr => m = 3.4×59 => m = 200.6 gr