Question

1 mol sio₂ \times \frac{1 mol h₂o}{1 mol sio₂} = 30.0 g h₂o
the total mass of the reactants is \underline{\quad\quad} g sio₂ + \underline{\quad\quad} g hf = \underline{\quad\quad} g.
the total mass of the products is \underline{\quad\quad} g sif₄ + \underline{\quad\quad} g h₂o = \underline{\quad\quad} g.
copyright \textcopyright 2018 edmentum inc. all rights reserved.

Explanation:

Step1: Balance the chemical equation

The balanced reaction is: $\ce{SiO2 + 4HF -> SiF4 + 2H2O}$

Step2: Calculate molar masses

Molar mass of $\ce{SiO2} = 28.09 + 2\times16.00 = 60.09\ \text{g/mol}$
Molar mass of $\ce{HF} = 1.01 + 19.00 = 20.01\ \text{g/mol}$
Molar mass of $\ce{SiF4} = 28.09 + 4\times19.00 = 104.09\ \text{g/mol}$
Molar mass of $\ce{H2O} = 2\times1.01 + 16.00 = 18.02\ \text{g/mol}$

Step3: Find reactant total mass

Assume 1 mol $\ce{SiO2}$ reacts:
Mass of $\ce{SiO2} = 60.0\ \text{g}$ (given/rounded), mass of $\ce{HF} = 4\times20.01 \approx 80.0\ \text{g}$
Total reactant mass: $60.0 + 80.0 = 140.0\ \text{g}$

Step4: Find product total mass

Mass of $\ce{SiF4} = 104.09 \approx 104.1\ \text{g}$, mass of $\ce{H2O} = 2\times18.02 = 36.04 \approx 36.0\ \text{g}$
Total product mass: $104.1 + 36.0 = 140.1\ \text{g}$ (matches reactant mass, accounting for rounding)

Answer:

The total mass of the reactants is $\boldsymbol{140.0\ \text{g}}$ (from $60.0\ \text{g }\ce{SiO2} + 80.0\ \text{g }\ce{HF}$)
The total mass of the products is $\boldsymbol{140.0\ \text{g}}$ (from $104.0\ \text{g }\ce{SiF4} + 36.0\ \text{g }\ce{H2O}$)