The molar mass of the water insoluble product formed from the fusion of chromite ore \(FeCr_2\text{O}_4\) with \(Na_2\text{CO}_3\) in presence of \(O_2\) is ....... g mol\(^{-1}\):
To solve this problem, we need to identify the insoluble product formed in the reaction and then calculate its molar mass.
1. Identifying the Reaction:
Chromite ore ($FeCr_2O_4$) reacts with sodium carbonate ($Na_2CO_3$) and oxygen ($O_2$) to form sodium chromate ($Na_2CrO_4$), iron(III) oxide ($Fe_2O_3$), and carbon dioxide ($CO_2$).
The balanced chemical equation is:
$4FeCr_2O_4(s) + 8Na_2CO_3(s) + O_2(g) \rightarrow 8Na_2CrO_4(aq) + 2Fe_2O_3(s) + 8CO_2(g)$
2. Identifying the Insoluble Product:
From the reaction, sodium chromate is soluble, while iron(III) oxide ($Fe_2O_3$) is insoluble.
3. Calculating the Molar Mass of $Fe_2O_3$:
Given:
Molar mass of Fe = 56 g/mol
Molar mass of O = 16 g/mol
Molar mass of $Fe_2O_3$ = (2 × Molar mass of Fe) + (3 × Molar mass of O) = (2 × 56 g/mol) + (3 × 16 g/mol) = 112 g/mol + 48 g/mol = 160 g/mol
Final Answer:
The molar mass of the insoluble product $Fe_2O_3$ is 160 g/mol.
Chromite ore, with the formula FeCr₂O₄, contains iron (Fe) and chromium (Cr). When it’s heated (fused) with sodium carbonate (Na₂CO₃) and oxygen (O₂), a chemical reaction occurs. This is a common process to extract chromium from its ore.
The balanced chemical equation for the reaction is:
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 2Fe₂O₃ + 8NaCrO₂ + 8CO₂
Reactants: FeCr₂O₄ (chromite ore), Na₂CO₃ (sodium carbonate), O₂ (oxygen).
Products:
The question asks for the water-insoluble product. Let’s analyze the products:
Therefore, the water-insoluble product is Fe₂O₃.
The molar mass of a compound is the sum of the atomic masses of its atoms. Let’s calculate for Fe₂O₃:
Atomic masses:
Fe₂O₃ has 2 iron atoms and 3 oxygen atoms:
Total molar mass of Fe₂O₃ = 111.70 + 48.00 = 159.70 g/mol
The molar mass of the water-insoluble product (Fe₂O₃) is 159.70 g/mol.
Among $ 10^{-10} $ g (each) of the following elements, which one will have the highest number of atoms?
Element : Pb, Po, Pr and Pt
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
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Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to: