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How many grams of calcium nitrate would react with 4.55 moles of Chromium (III) sulfate to produce calcium sulfate and chromium (III) nitrate?

A.) 6.86x10^2 g
B.) 3.43x10^2 g
C.) 6.04x10^-2 g
D.) 3.09x10^3 g

Respuesta :

The equation of the reaction is;
3Ca(NO3)2 + Cr2(SO4)3 = 3Ca(SO4) + 2 Cr(NO3)3
4.55 moles of chromium (iii) sulfate reacts with calcium nitrate
The mole ratio of Chromium (iii) sulfate to calcium nitrate is 1: 3
Therefore; the moles of calcium nitrate will be given by 
  = (4.55/ 1)×3
 = 13.65 moles
Hence; 13.65 moles of calcium nitrate would react with 4.55 moles of Chromium (iii) sulfate.

The mass in grams of calcium nitrate would react with 4.55 moles of Chromium (III) sulfate to produce calcium sulfate and chromium (III) nitrate is 2.2 × 10³g.

How do we calculate mass from moles?

Mass of any substance will be calculated by using the moles as:

n = W/M ,where

W = required mass

M = molar mass

Given chemical reaction is:

3Ca(NO₃)₂ + Cr₂(SO₄)₃ → 3Ca(SO₄) + 2Cr(NO₃)₃

From the stoichiometry of the reaction, it is clear that:

1 mole of Cr₂(SO₄)₃ = react with 3 moles of Ca(NO₃)₂

4.55 moles of Cr₂(SO₄)₃ = react with 3×4.55=13.65 moles of Ca(NO₃)₂

Now mass of Ca(NO₃)₂ will be calculated as:
W = (13.65)(164.08) = 2239.69 g = 2.2 × 10³g

Hence, required mass of Ca(NO₃)₂ is 2239.69g.

To know more about moles, visit the below link:

https://brainly.com/question/24631381

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