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  1. 8.702g of HCl

    Explanation:

    Using the details from the equation;

    4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)

    The molecular mass of HCl is; 4 (1 + 35.5) = 146g

    The molecular mass of MnO2 is; (54.9 + 32) = 86.9g

    From the equation it means;

    86.9g of MnO2 reacts with 146g of HCl

    if 1 g of MnO2  reacts with  146 ÷ 86.9 of HCl

    ∴ 5.18g of MnO2 will react with; 146 ÷ 86.9 x  5.18 = 8.702g of HCl

    Note: Approximations were made during the determination of the molecular masses.

  2. [tex]m_{Mg}=30.8mgMg[/tex]

    Explanation:

    Hello,

    Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

    [tex]n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2[/tex]

    Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

    [tex]m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg[/tex]

    Best regards.

  3. The mass of bromine reacted is 160.6 grams.

    Explanation:

    To calculate the number of moles, we use the equation:

    [tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]        .....(1)

    Given mass of aluminium = 18.1 g

    Molar mass of aluminium = 27 g/mol

    Putting values in equation 1, we get:

    [tex]\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol[/tex]

    The chemical equation for the reaction of aluminium and bromide follows:

    [tex]2Al+3Br_2\rightarrow 2AlBr_3[/tex]

    By Stoichiometry of the reaction:

    2 moles of aluminium reacts with 3 moles of bromine gas

    So, 0.670 moles of aluminium will react with = [tex]\frac{3}{2}\times 0.670=1.005mol[/tex] of bromine gas.

    Now, calculating the mass of bromine gas, we use equation 1:

    Moles of bromine gas = 1.005 moles

    Molar mass of bromine gas = 159.81 g/mol

    Putting values in equation 1, we get:

    [tex]1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g[/tex]

    Hence, the mass of bromine reacted is 160.6 grams.

  4. Answer : The mass of [tex]Mg[/tex] required is 30.38 mg.

    Explanation :

    To calculate the moles of hydrogen gas, we use the equation given by ideal gas :

    PV = nRT

    where,

    P = Pressure of hydrogen gas = 754 torr

    V = Volume of the hydrogen gas = 31.2 mL = 0.0312 L

    n = number of moles of hydrogen gas = ?

    R = Gas constant = [tex]62.364\text{ L.torr }mol^{-1}K^{-1}[/tex]

    T = Temperature of hydrogen gas = [tex]25^oC=273+25=298K[/tex]

    Putting values in above equation, we get:

    [tex]754torr\times 0.0312L=n\times 62.364\text{ L.torr }mol^{-1}K^{-1}\times 298K\\\\n=1.266\times 10^{-3}mole[/tex]

    Now we have to calculate the moles of [tex]Mg[/tex].

    The balanced chemical reaction is:

    [tex]Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)[/tex]

    From the balanced chemical reaction, we conclude that

    As, 1 mole of [tex]H_2[/tex] produced from 1 mole [tex]Mg[/tex]

    As, [tex]1.266\times 10^{-3}[/tex] mole of [tex]H_2[/tex] produced from [tex]1.266\times 10^{-3}[/tex] mole [tex]Mg[/tex]

    Now we have to calculate the mass of [tex]Mg[/tex].

    Molar mass of [tex]Mg[/tex] = 24 g/mol

    [tex]\text{Mass of }Mg=\text{Moles of }Mg\times \text{Molar mass of }Mg[/tex]

    [tex]\text{Mass of }Mg=1.266\times 10^{-3}mole\times 24g/mole=30.38\times 10^{-3}g=30.38mg[/tex]

    conversion used : (1 g = 1000 mg)

    Therefore, the mass of [tex]Mg[/tex] required is 30.38 mg

  5. 1.1g of copper

    Explanation: We begin by writing the balanced reaction equation.

    So we have : Cu(s) + 2AgNO3 (aq) > Cu(NO3)2 (aq) + 2Ag (s)

    Molar mass of AgNO3 = 107.87 + 14.01 + 3(16.0) = 169.88g/mol, for 2 moles we then have 2 x 169.88 =339.76g.

    Atomic mass of Copper = 64g approximately

    From the equation the following deductions can be made:

    339.76g of AgNO3 reacts with 64g of copper

    5.65g of AgNO3 would react with 64/ 339.76 x 5.65 =1.062 approx 1.1g of copper.

  6. Answer : 70.906 grams of chlorine reacted with hydrogen

    Explanation :

    Step 1 : Write balanced chemical equation.

    The balanced chemical equation for the reaction between hydrogen and chlorine gas is given below.

    [tex]H_{2} (g) + Cl_{2} (g) \rightarrow 2HCl(g)[/tex]

    Step 2 : Find moles of H₂ gas.

    The moles of H₂ can be found as

    [tex]mole = \frac{grams}{MolarMass}[/tex]

    We have 2.0200 g of H₂ and molar mass of H₂ is 2.02 g/mol.

    Let us plug in these values to find moles of H₂.

    [tex]mole =\frac{2.0200g}{2.02g/mol} = 1 mol[/tex]

    We have 1 mol of H₂.

    Step 3 : Find moles of Cl₂ using mole ratio.

    The mole ratio of H₂ and Cl₂ is 1 : 1.

    The moles of Cl₂ can be calculated as

    [tex]1 mol H_{2} \times\frac{1 mol Cl_{2}}{1 mol H_{2}} = 1 mol Cl_{2}[/tex]

    Step 4 : Find grams of Cl₂.

    Molar mass of Cl₂ gas is 70.096 g/mol

    Mass of Cl₂ = [tex]1 mol Cl_{2} \times\frac{70.906g}{mol} = 70.906 g[/tex]

    We have 70.906 grams of Cl₂

  7. 3.89 g Mg

    Explanation:

    2M = 2 mol/L

    120 mL = 0.120 L

    2 mol/L *0.120 L = 0.240 mol H3PO4

                                   3Mg + 2H3PO4 > Mg3(PO4)2 + 3H2

    from reaction         3 mol   2 mol

    given                  0.240mol  x mol

    x = 0.240*2/3 = 0.16 mol Mg

    M(Mg) = 24.3 g/mol

    24.3 g/mol *0.16 mol = 3.89 g Mg  

  8. a) Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

    b) 2.40 moles HF

    c) 5.25 grams NaF

    d)0.609 grams Na2SiO3

    e) 0.89 grams Na2SiO3

    Explanation:

    Step 1: Data given

    Step 2: The balanced equation

    Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

    b) How many moles of HF are needed to react with 0.300 mol of Na2SiO3?

    For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

    For 0.300 moles Na2SiO3 we'll need 8*0.300 = 2.40 moles HF

    c. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?

    For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

    For 0.500 moles we'll have 0.500 / 4 = 0.125 moles NaF

    Mass NaF = 0.125 moles * 41.99 g/mol

    Mass NaF = 5.25 grams NaF

    d. How many grams of Na2SiO3 can react with 0.800 g of HF?

    Moles HF = 0.800 grams / 20.01 g/mol

    Moles HF = 0.0399 moles

    For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

    For 8 moles we need 1 moles Na2SiO3 to react

    For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

    Mass Na2SiO3 = 0.00499 moles * 122.06 g/mol  

    Mass Na2SiO3 = 0.609 grams Na2SiO3

    Using your answer to part d-If you started with 1.5 g of Na2SiO3how many grams would be left after the reaction is complete?

    Number of moles HF = 0.0399 moles

    Number of moles Na2SiO3 = 1.5 grams / 122.06 g/mol = 0.0123 moles

    For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

    For 8 moles we need 1 moles Na2SiO3 to react

    For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

    There will remain 0.0123 - 0.00499 = 0.00731 moles  Na2SiO3

    Mass Na2SiO3 remaining = 0.00731 * 122.06 g/mol = 0.89 grams Na2SiO3

  9. 30.7 mg of Magnesium

    Explanation: The equation of reaction is;

    Mg(s) + 2HCl(aq) > MgCl2(aq) + H2(g).

    I mol of Magnessium produces 1 mol of Hydrogen gas.

    Pressure,p= 754 torr, volume, v= 31.2 ml( which is also equals to 0.0312L) , R= 62.36367L torr/kmol, temperature, T = 25 + 273.15)

    Step 1: find the number of moles using PV = nRT.

    Making n(number of mole) the subject of the formula, we have;

    n = PV/RT

    Substitute the above parameters into the formula, we have;

    n= (754 Torr) x (0.0312 L) / ((62.36367 L torr/K mol) x (25.0 + 273.15) K) = 0.001265 mol of Hydrogen gas.

    Step 2: using n = mass/molar mass.

    Mass = n × molar mass

    (0.001265 mol of H2) x (1 mol Mg / 1 mol H2) x (24.30506 g Mg/mol) = 0.030745 g = 30.7 mg of Magnessium

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