How many grams of ZrCl4 can be produced if 123 g of ZrSiO4 react with 85.0 g of Cl2?

153.3 grams of ZrCl₄ are produced

Explanation:

The equation of the reaction is as follows:

ZrSiO₄ + Cl₂ > ZrCl₄ + SiO₂ + O₂

molar mass of ZrSiO₄ = (91 + 32 + 16 * 4) = 187.0 g/mol

molar mass of ZrCl₄ = (91 + 35.5 * 4) = 233.0 g/mol

molar mass of Cl₂ = (35.5 * 2) 71.0 g/mol

From the equation of reaction, 1 mole of ZrSiO₄ reacts with one mole of  Cl₂ to produce one mole of ZrCl₄

number of moles of ZrSiO₄ present in 123 g = 123/187 = 0.65 moles

number of moles of Cl₂ present in 85.0 g = 85.0/71.0 = 1.19 moles

therefore, ZrSiO₄ is the limiting reactant

123.0 g of ZrSiO₄ will react with excess Cl₂ to produce 123 * 233/187 grams of ZrCl₄ = 153.3 grams of ZrCl₄

Therefore, 153.3 grams of ZrCl₄ are produced

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer