If a $49.9 \cdot g$ $49.9*g$ mass of sodium hydroxide reacts with sulfuric acid, what mass of sodium sulfate results?

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If a $49.9 \cdot g$ $49.9*g$ mass of sodium hydroxide reacts with sulfuric acid, what mass of sodium sulfate results?

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Answer 1 · 2017-03-24T17:13:14

If we assume $sodium hydroxide$ $"sodium hydroxide"$ reacts with $sulfuric acid$ $"sulfuric acid"$ .....approx. $90 \cdot g$ $90*g$ $sodium sulfate$ $"sodium sulfate"$ results......

Explanation:

We write the equation as:

$2 N a O H (a q) + H_{2} S O_{4} (a q) \to N a_{2} S O_{4} (a q) + 2 H_{2} O (l)$ $2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) + 2H_2O(l)$

$Moles of sodium hydroxide = \frac{49.9 \cdot g}{40.00 \cdot g \cdot m o l^{- 1}} = 1.25 \cdot m o l$ $"Moles of sodium hydroxide"=(49.9*g)/(40.00*g*mol^-1)=1.25*mol$ .

The equation specifies that HALF an equiv of $sodium sulfate$ $"sodium sulfate"$ results from neutralization, and this represents a mass of

$1.25 \cdot m o l \times \frac{1}{2} \times 142.04 \cdot g \cdot m o l^{- 1} = 88.8 \cdot g .$ $1.25*molxx1/2xx142.04*g*mol^-1=88.8*g.$

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If a $49.9 \cdot g$ $49.9*g$ mass of sodium hydroxide reacts with sulfuric acid, what mass of sodium sulfate results?

1 Answer

Explanation:

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Math

Humanities

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If a 49.9⋅g49.9*g mass of sodium hydroxide reacts with sulfuric acid, what mass of sodium sulfate results?

1 Answer

Explanation:

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If a $49.9 \cdot g$ $49.9*g$ mass of sodium hydroxide reacts with sulfuric acid, what mass of sodium sulfate results?