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Dmitriy789 [7]
3 years ago
10

Level 1: 2Sr + 02 > 2Sro

Chemistry
1 answer:
tekilochka [14]3 years ago
8 0

Answer:

Sr would be the limiting reactant

5 moles

Explanation:

Since the equation is a balanced equation, the coefficient shows how each substance relates to the other in terms of the number of moles.

Reactants would be those on the left hand side of the arrow, while the products would be found on te right and side of the arrow. In this question, the reactants would be Sr and O₂.

Limiting reactant is the reactant that is insufficient; meaning to say that there is not enough of that substance and thus the reaction cannot continue. The other reactant(s) that is not limiting is called the excess reactants.

From the balanced equation, 2 moles of Sr is needed to react with 1 mole of O₂. Thus, if we have 5 moles of each reactant, Sr would be the limiting reactant since for every 1 mole of O₂, there has to be 2 moles of Sr in order for the reaction to proceed. Thus, if we have 5 moles of O₂, we would need 10 moles of Sr.

When we work out the amount of products formed, we look at the number of moles of the limiting reactant. This is because the limiting reactant determines how much is being reacted, while the excess number of moles of the excess reactant will remain unreacted.

For every 2 moles of Sr reacted, 2 moles of SrO would be produced. This means that the mole ratio of Sr to SrO is 1:1. Thus, since 5 moles of Sr has been reacted, 5 moles of the product (SrO) would be produced.

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3 years ago
Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati
ollegr [7]

Explanation:

Let us assume that the given data is as follows.

   mass of barium acetate = 2.19 g

   volume = 150 ml = 0.150 L    (as 1 L = 1000 ml)

   concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

        Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}

Hence, moles of C_{2}H_{3}O^{-}_{2} = 2 \times Ba(C_{2}H_{3}O_{2})_{2}  .......... (1)

As,   No. of moles = \frac{mass}{\text{molar mass}}

Hence, moles of Ba(C_{2}H_{3}O_{2})_{2} will be calculated as follows.                          

     No. of moles = \frac{mass}{\text{molar mass}}  

                          =  \frac{2.19 g}{255.415 g/mol}   (molar mass of Ba(C_{2}H_{3}O_{2})_{2} is 255.415 g/mol)            

                       = 8.57 \times 10^{-3}

    Moles of C_{2}H_{3}O^{-}_{2} = 2 \times 8.57 \times 10^{-3}

                          = 0.01715 mol

Hence, final molarity will be as follows.

              Molarity = \frac{\text{no. of moles}}{volume}

                             = \frac{0.01715 mol}{0.150 L}

                             = 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

5 0
3 years ago
How many liters would 4.56 mol sulfur dioxide occupy at STP?
algol13

The volume of Sulfur dioxide-SO₂ at STP : = 102.144 L

<h3>Further explanation</h3>

Given

4.56 mol Sulfur dioxide-SO₂

Required

The volume

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.

so for 4.56 mol :

= 4.56 x 22.4 L

= 102.144 L

4 0
3 years ago
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Answer:

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Explanation:

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=\dfrac{1.0505\times 10^{23}}{6.022\times 10^{23}}\times 44\\\\=7.675\ \text{grams}

  • Hence, 7.675 grams of carbon dioxide is present in 1.505\times 10^{23} molecules of the gas.
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3 years ago
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