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blagie [28]
3 years ago
5

If 1.0g of rubidium and 1.0 g of bromine are reacted what will be left in measurable amounts in the reaction

Chemistry
1 answer:
pentagon [3]3 years ago
7 0

Answer:

                     0.0639 g of Br₂ will be left

Explanation:

                    The balance chemical equation for given synthesis reaction is,

                                          2 Rb + Br₂ → 2 RbBr

Step 1: <u>Calculate Moles of each reactant for given masses as;</u>

Moles of Rb = 1 g ÷ 85.46 g/mol

Moles of Rb = 0.0117 moles

Similarly,

Moles of Br₂ = 1 g ÷ 159.80 g/mol

Moles of Br₂ = 0.00625 moles

Step 2: <u>Find reactant in Excess as:</u>

According to equation,

                    2 moles of Rb reacted with  =  1 mole of Br₂

So,

              0.0117 moles of Rb will react with  =  X moles of Br₂

Solving for X,

                      X  =  0.0117 mol × 1 mol / 2 mol

                     X =  0.00585 mol of Br₂

While, as calculated above we are provided with 0.00625 moles of Br₂. Therefore, Br₂ is in excess and the excess amount is calculated as,

= Given Moles - Consumed Moles

= 0.00625 moles - 0.00585 moles

= 0.0004 moles

Step 3: <u>Calculate mass of excess Br₂ as:</u>

Mass = Moles × M.Mass

Mass = 0.0004 mol × 159.80

Mass = 0.0639 g of Br₂

Note: Speaking broadly, there will be 1.93 g of RbBr along with 0.0639 g of Br₂ because RbBr is being produced and is available along with unreacted Br₂.

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The sample of argon gas that has the same number of atoms as a 100 milliliter sample of helium gas at 1.0 atm and 300 is 100. mL at 1.0 atm and 300. K

The correct option is D.

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The number of moles of gases in each of the given samples of gas is found below using the ideal gas equation.

The ideal gas equation is: PV/RT = n

where;

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n =  1 * 0.1 / 0.082 * 300

n = 0.00406 moles

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A. n =  1 * 0.05 / 0.082 * 300

n = 0.00203 moles

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n = 0.00203 moles

D. n =  1 * 0.1 / 0.082 * 300

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Learn more about ideal gas equation at: brainly.com/question/24236411

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