You are reacting hydrogen gas with tin oxide to purify tin. SnO2 + 2 H2 → Sn + 2 H2O. You have 45.8L of H2 gas and 351.3g of SnO2. What will

Question

You are reacting hydrogen gas with tin oxide to purify tin. SnO2 + 2 H2 → Sn + 2 H2O. You have 45.8L of H2 gas and 351.3g of SnO2. What will be the limiting reactant, and how much of the excess reactant will be left over?

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Hải Đăng 11 mins 2021-07-22T15:08:18+00:00 1 Answers 0 views 0

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    2021-07-22T15:10:15+00:00

    Answer: The limiting reactant is hydrogen gas and the mass of excess reactant (SnO_2) left over is 197.43 g

    Explanation:

    Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

    Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

    The number of moles is defined as the ratio of the mass of a substance to its molar mass.

    The equation used is:

    ……(1)

    We are given:

    Given mass of SnO_2 = 351.3 g

    Molar mass of SnO_2 = 150.71 g/mol

    Putting values in equation 1, we get:

    \text{Moles of }SnO_2=\frac{351.3g}{150.71g/mol}=2.33mol

    At STP conditions:

    22.4 L of volume is occupied by 1 mole of a gas

    Applying unitary method:

    45.8 L of volume will be occupied by = \frac{1mol}{22.4L}\times 45.8L=2.04mol of hydrogen gas

    For the given chemical reaction:

    SnO_2+2H_2\rightarrow Sn+2H_2O

    By stoichiometry of the reaction:

    If 2 moles of hydrogen gas reacts with 1 mole of SnO_2

    So, 2.04 moles of hydrogen gas will react with = \frac{1}{2}\times 2.04=1.02mol of SnO_2

    As the given amount of SnO_2 is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

    Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.

    Moles of excess reactant (SnO_2) left = [2.33 – 1.02] = 1.31 moles

    We know, molar mass of SnO_2 = 150.71 g/mol

    Putting values in equation 1, we get:

    \text{Mass of }SnO_2=(1.31mol\times 150.71g/mol)=197.43g

    Hence, the limiting reactant is hydrogen gas and the mass of excess reactant (SnO_2) left over is 197.43 g

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