Question

1. how many grams are in 1.11 moles of manganese sulfate, $\text{mn}_3(\text{so}_4)_7$?
2. how many moles are in 4.5 grams of sodium fluoride, $\text{naf}$?

Explanation:

Step1: Calculate molar mass of $\text{Mn}_3(\text{SO}_4)_7$

Molar mass of Mn = 54.94 g/mol, S = 32.07 g/mol, O = 16.00 g/mol

$$\begin{align*} M &= 3\times54.94 + 7\times(32.07 + 4\times16.00)\\ &= 164.82 + 7\times(32.07 + 64.00)\\ &= 164.82 + 7\times96.07\\ &= 164.82 + 672.49\\ &= 837.31\ \text{g/mol} \end{align*}$$

Step2: Convert moles to grams (Q6)

Use formula $m = n\times M$

$$ m = 1.11\ \text{mol} \times 837.31\ \text{g/mol} = 929.41\ \text{g} $$

Step3: Calculate molar mass of $\text{NaF}$

Molar mass of Na = 22.99 g/mol, F = 19.00 g/mol

$$ M = 22.99 + 19.00 = 41.99\ \text{g/mol} $$

Step4: Convert grams to moles (Q7)

Use formula $n = \frac{m}{M}$

$$ n = \frac{4.5\ \text{g}}{41.99\ \text{g/mol}} \approx 0.107\ \text{mol} $$

Answer:

1. 929.41 grams
2. 0.11 moles (rounded to two significant figures)