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Determine how many grams of precipitate (solid) will form when 621 g of magnesium
chloride react with excess silver nitrate in the reaction
2 AgNO3(aq) + MgCl(aq) - 2 AgCl(s) + Mg(NO3)2(aq)
NB: Convert to moles, then use your preferred method to get moles AgCl produced
from the given moles of MgCl, then convert moles AgCl produced to grams AgCI.

Sagot :

DragonPup315idn

Answer:

2,981g

Explanation:

Firstly, we need to find the number of moles of MgCl that we have by using the formula: mass = No. Moles x Molar Mass, which we can rearrange so that we are solving for no. moles:

No. Moles = mass / Molar Mass

We are given a mass of 621g, and we can calculate the molar mass of MgCl by adding the two molar masses together: 24.31+35.45 = 59.76

Now we can calculate number of moles by substituting these values into the formula:

n = 621 / 59.76

No. moles = 10.4

Now we can use the co-efficients in the formula to tell us how many moles of AgCl will be formed. The coefficient of MgCl is 1, and the coefficient of AgCl is 2. This means that every 1 mol of MgCl will form 2 moles of AgCl. So, to find the no. moles of AgCl, we multiply our no. moles by 2:

10.4 x 2 = 20.8 moles

Finally we convert this back into mass by multiplying the no. moles by the Molar mass of AgCl:

m = 20.8 x (107.87+35.45)

m = 2,981g

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